NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 13: Statistics and Probability

Exercise 13.3

Step 1: Find the class marks (x).

The class mark is the middle point of each class interval.

• For 1–3: \(\dfrac{1+3}{2} = 2\)
• For 3–5: \(\dfrac{3+5}{2} = 4\)
• For 5–7: \(\dfrac{5+7}{2} = 6\)
• For 7–10: \(\dfrac{7+10}{2} = 8.5\)

Step 2: Make a table of f × x.

Step 3: Use the formula of mean.

The formula is: \(\bar{x} = \dfrac{\sum f x}{\sum f}\)

Here, \(\sum f = 75\), and \(\sum f x = 412.5\).

Step 4: Put the values.

\(\bar{x} = \dfrac{412.5}{75} = 5.5\)

Final Answer: The mean of the distribution is 5.5.

Step 1: Write down the class intervals and frequencies.

We are given:

• Marks (class intervals): 10–20, 20–30, 30–40, 40–50, 50–60
• Number of students (frequency, f): 2, 4, 7, 6, 1

Step 2: Find the class marks (x).

Class mark = (lower limit + upper limit) ÷ 2

• For 10–20: (10 + 20) ÷ 2 = 15
• For 20–30: (20 + 30) ÷ 2 = 25
• For 30–40: (30 + 40) ÷ 2 = 35
• For 40–50: (40 + 50) ÷ 2 = 45
• For 50–60: (50 + 60) ÷ 2 = 55

So, class marks are: 15, 25, 35, 45, 55.

Step 3: Multiply frequency (f) with class mark (x).

Step 4: Add the totals.

• Total of frequencies, \(\sum f = 2 + 4 + 7 + 6 + 1 = 20\)
• Total of f × x, \(\sum f x = 30 + 100 + 245 + 270 + 55 = 700\)

Step 5: Apply the formula of mean.

Mean, \(\bar{x} = \dfrac{\sum f x}{\sum f}\)

\(\bar{x} = \dfrac{700}{20} = 35\)

Final Answer: The mean marks = 35.

Step 1: Find the class marks (mid-points)

The class mark of each class is the middle value. It is found by the formula:

\( \text{Class mark} = \dfrac{\text{Lower limit + Upper limit}}{2} \)

For 4–7: \( (4+7)/2 = 5.5 \)

For 8–11: \( (8+11)/2 = 9.5 \)

For 12–15: \( (12+15)/2 = 13.5 \)

For 16–19: \( (16+19)/2 = 17.5 \)

Class marks: 5.5, 9.5, 13.5, 17.5

Step 2: Multiply frequency (f) with class mark (x)

Now add the totals:

\( \sum f = 5+4+9+10 = 28 \)

\( \sum f x = 27.5+38+121.5+175 = 362 \)

Step 3: Apply the formula for mean

Formula: \( \bar{x} = \dfrac{\sum f x}{\sum f} \)

\( \bar{x} = \dfrac{362}{28} \)

Step 4: Simplify

\( \bar{x} = 12.93 \) (approx.)

Final Answer: Mean ≈ 12.93

Step 1: Write class intervals and their frequencies

Class intervals (pages/day): 16–18, 19–21, 22–24, 25–27, 28–30

Number of days (frequency, f): 1, 3, 4, 9, 13

Step 2: Find the mid-point (x) of each class interval

• (16 + 18) ÷ 2 = 17
• (19 + 21) ÷ 2 = 20
• (22 + 24) ÷ 2 = 23
• (25 + 27) ÷ 2 = 26
• (28 + 30) ÷ 2 = 29

So mid-points are: 17, 20, 23, 26, 29

Step 3: Multiply frequency (f) by mid-point (x)

Step 4: Find totals

Σf = 1 + 3 + 4 + 9 + 13 = 30

Σ(f × x) = 17 + 60 + 92 + 234 + 377 = 780

Step 5: Apply mean formula

Mean = Σ(f × x) ÷ Σf

= 780 ÷ 30

= 26 pages/day

Final Answer: 26 pages/day

Step 1: Write the class intervals and their frequencies (f).

Step 2: Find the class marks (mid-points).

Formula: \(x = \frac{\text{lower limit} + \text{upper limit}}{2}\)

• For 1–200: \(x = \frac{1 + 200}{2} = 100.5\)
• For 201–400: \(x = \frac{201 + 400}{2} = 300.5\)
• For 401–600: \(x = \frac{401 + 600}{2} = 500.5\)
• For 601–800: \(x = \frac{601 + 800}{2} = 700.5\)

Step 3: Multiply frequency (f) with class mark (x).

Step 4: Find totals.

• \(\sum f = 14 + 15 + 14 + 7 = 50\)
• \(\sum (f × x) = 17825\)

Step 5: Apply the formula for Mean.

\(\bar{x} = \frac{\sum f x}{\sum f}\)

\(\bar{x} = \frac{17825}{50} = 356.5\)

Final Answer: Mean daily income = Rs 356.5

Step 1: Identify the data.

We are given the number of seats occupied in groups (called classes) and how many times they occurred (called frequency).

Step 2: Find the mid-point (class mark) of each interval.

The class mark is the average of the lower and upper limits:

• For 100–104: \( \tfrac{100+104}{2} = 102 \)
• For 104–108: \( \tfrac{104+108}{2} = 106 \)
• For 108–112: \( \tfrac{108+112}{2} = 110 \)
• For 112–116: \( \tfrac{112+116}{2} = 114 \)
• For 116–120: \( \tfrac{116+120}{2} = 118 \)

Step 3: Multiply frequency by class mark.

We calculate \( f \times x \):

Step 4: Add the values.

• Total frequency: \( \sum f = 15+20+32+18+15 = 100 \)
• Total of \( f \times x \): \( \sum fx = 1530+2120+3520+2052+1770 = 10992 \)

Step 5: Apply the mean formula.

The mean is given by:

\[ \bar{x} = \frac{\sum f x}{\sum f} \]

Substituting the values:

\[ \bar{x} = \frac{10992}{100} = 109.92 \]

Final Answer: The mean number of seats occupied = 109.92 seats.

Step 1: Identify the class intervals and their frequencies.

The weights are divided into groups (called class intervals):

• 100–110 (4 wrestlers)
• 110–120 (14 wrestlers)
• 120–130 (21 wrestlers)
• 130–140 (8 wrestlers)
• 140–150 (3 wrestlers)

Step 2: Find the class marks (midpoints).

For each class interval, the class mark = (lower limit + upper limit) ÷ 2.

• For 100–110: (100 + 110) ÷ 2 = 105
• For 110–120: (110 + 120) ÷ 2 = 115
• For 120–130: (120 + 130) ÷ 2 = 125
• For 130–140: (130 + 140) ÷ 2 = 135
• For 140–150: (140 + 150) ÷ 2 = 145

Step 3: Multiply each frequency (f) with its class mark (x).

Step 4: Find the totals.

Total frequency (Σf) = 4 + 14 + 21 + 8 + 3 = 50

Total of f × x (Σfx) = 420 + 1610 + 2625 + 1080 + 435 = 6170

Step 5: Apply the mean formula.

Mean (\(\bar{x}\)) = Σfx ÷ Σf

\(\bar{x} = \dfrac{6170}{50} = 123.4\)

Final Answer: The mean weight of the wrestlers is 123.4 kg.

Step 1: Identify the data.

• The table shows classes of mileage (in km per litre) and the number of cars in each class.
• Total number of cars = 50.

Step 2: Find the class marks (mid-points).

For each class interval, the class mark = (lower limit + upper limit) ÷ 2.

• For 10–12: (10 + 12) ÷ 2 = 11
• For 12–14: (12 + 14) ÷ 2 = 13
• For 14–16: (14 + 16) ÷ 2 = 15
• For 16–18: (16 + 18) ÷ 2 = 17

Step 3: Multiply class mark with frequency (number of cars).

Step 4: Apply the formula for Mean.

Mean mileage (\( \bar{x} \)) = (Σ f × x) ÷ (Σ f)

= 724 ÷ 50

= 14.48 km/l

Step 5: Compare with manufacturer’s claim.

The manufacturer claimed the mileage = 16 km/l.

But the actual mean mileage = 14.48 km/l.

Conclusion: The claim of 16 km/l is not supported by the data.

Step 1: Look at the given table. The weights are divided into groups (called class intervals) like 40–45, 45–50, 50–55, etc. The number of persons in each group is given below it.

Step 2: In a less-than cumulative frequency table, we keep adding the frequencies step by step. For each class, we take "less than upper value".

Step 3: Start with the first class (40–45). - Upper value = 45 - No. of persons in this class = 4 - So, less than 45 → cumulative frequency = 4.

Step 4: Next class (45–50). - Upper value = 50 - Persons in this class = 4 - Add to previous cumulative (4 + 4 = 8) - So, less than 50 → cumulative frequency = 8.

Step 5: Next class (50–55). - Upper value = 55 - Persons in this class = 13 - Add to previous cumulative (8 + 13 = 21) - So, less than 55 → cumulative frequency = 21.

Step 6: Next class (55–60). - Upper value = 60 - Persons in this class = 5 - Add to previous cumulative (21 + 5 = 26) - So, less than 60 → cumulative frequency = 26.

Step 7: Next class (60–65). - Upper value = 65 - Persons in this class = 6 - Add to previous cumulative (26 + 6 = 32) - So, less than 65 → cumulative frequency = 32.

Step 8: Next class (65–70). - Upper value = 70 - Persons in this class = 5 - Add to previous cumulative (32 + 5 = 37) - So, less than 70 → cumulative frequency = 37.

Step 9: Next class (70–75). - Upper value = 75 - Persons in this class = 2 - Add to previous cumulative (37 + 2 = 39) - So, less than 75 → cumulative frequency = 39.

Step 10: Last class (75–80). - Upper value = 80 - Persons in this class = 1 - Add to previous cumulative (39 + 1 = 40) - So, less than 80 → cumulative frequency = 40.

Step 11: Write all values in a new table as shown in the answer. This is the less-than type cumulative frequency table.

Step 1: Read the given data carefully. The table shows cumulative frequencies (less–than type). For example, “Below 30 = 130” means that 130 students scored less than 30 marks.

Step 2: To convert cumulative frequencies into ordinary frequencies, we need to subtract consecutive values. Why? Because cumulative means “total so far”. Ordinary frequency means “number of students in that specific interval”.

Step 3: Start with the first class (0–10). - Below 10 = 10. - This is the first cumulative value, so frequency for 0–10 = 10.

Step 4: Now calculate the second class (10–20). - Below 20 = 50. - Below 10 = 10. - Frequency for 10–20 = 50 – 10 = 40.

Step 5: For the third class (20–30): - Below 30 = 130. - Below 20 = 50. - Frequency = 130 – 50 = 80.

Step 6: For the fourth class (30–40): - Below 40 = 270. - Below 30 = 130. - Frequency = 270 – 130 = 140.

Step 7: For the fifth class (40–50): - Below 50 = 440. - Below 40 = 270. - Frequency = 440 – 270 = 170.

Step 8: For the sixth class (50–60): - Below 60 = 570. - Below 50 = 440. - Frequency = 570 – 440 = 130.

Step 9: For the seventh class (60–70): - Below 70 = 670. - Below 60 = 570. - Frequency = 670 – 570 = 100.

Step 10: For the eighth class (70–80): - Below 80 = 740. - Below 70 = 670. - Frequency = 740 – 670 = 70.

Step 11: For the ninth class (80–90): - Below 90 = 780. - Below 80 = 740. - Frequency = 780 – 740 = 40.

Step 12: For the tenth class (90–100): - Below 100 = 800. - Below 90 = 780. - Frequency = 800 – 780 = 20.

Step 13: Now write the ordinary frequency distribution table:

Final Check: Add all frequencies: 10 + 40 + 80 + 140 + 170 + 130 + 100 + 70 + 40 + 20 = 800. This matches the total number of students given. ✔️

Step 1: Look at the given table. It is in “more than or equal to” form. This means: for each mark limit, the number tells us how many students scored at least those marks.

Step 2: To convert into a normal frequency table (0–10, 10–20, etc.), we must subtract successive values.

Step 3: Calculate frequencies one by one.

• For 0–10: Total candidates (≥0) – candidates (≥10) = 34 – 32 = 2
• For 10–20: Candidates (≥10) – candidates (≥20) = 32 – 30 = 2
• For 20–30: Candidates (≥20) – candidates (≥30) = 30 – 27 = 3
• For 30–40: Candidates (≥30) – candidates (≥40) = 27 – 23 = 4
• For 40–50: Candidates (≥40) – candidates (≥50) = 23 – 17 = 6
• For 50–60: Candidates (≥50) – candidates (≥60) = 17 – 11 = 6
• For 60–70: Candidates (≥60) – candidates (≥70) = 11 – 6 = 5
• For 70–80: Candidates (≥70) – candidates (≥80) = 6 – 4 = 2
• For 80–90: Candidates (≥80) – candidates (≥90). Since ≥90 is not given, we assume it is 0. So, 4 – 0 = 4

Step 4: Write all these frequencies in the table beside their class intervals.

Final Note: Always remember: when data is given in “more than or equal to” form, you can find the actual frequency by subtracting successive entries.

Step 1: Understand the terms

- Frequency: how many students are in that height group.

- Cumulative frequency: the total number of students from the first group up to the current group.

Step 2: First row

Frequency = 12. Since this is the first row, the cumulative frequency is the same:
\(a = 12\).

Step 3: Second row

Cumulative frequency given = 25. This means:
(First frequency) + (Second frequency) = 25.
So, \(12 + b = 25 \implies b = 13\).

Step 4: Third row

Frequency = 10. The cumulative frequency = previous cumulative (25) + 10 = 35.
So, \(c = 35\).

Step 5: Fourth row

Cumulative frequency given = 43. This means:
previous cumulative (35) + d = 43.
So, \(d = 43 - 35 = 8\).

Step 6: Fifth row

Cumulative frequency given = 48. This means:
previous cumulative (43) + e = 48.
So, \(e = 48 - 43 = 5\).

Step 7: Sixth row

Frequency = 2. The final cumulative frequency = previous cumulative (48) + 2 = 50.
So, \(f = 50\).

Final Answer: a = 12, b = 13, c = 35, d = 8, e = 5, f = 50.

Step 1: Understand the data.

The table gives us age groups (class intervals) of patients and how many patients are in each group (frequency).

Step 2: Less-than type cumulative frequency.

• For "less than 20": Only the first group (10–20) is included. So CF = 60.
• For "less than 30": Add patients in 10–20 and 20–30 → 60 + 42 = 102.
• For "less than 40": Add one more group → 102 + 55 = 157.
• For "less than 50": Add one more group → 157 + 70 = 227.
• For "less than 60": Add one more group → 227 + 53 = 280.
• For "less than 70": Add the last group → 280 + 20 = 300.

Step 3: More-than type cumulative frequency.

• Start with total patients = 300.
• For "more than or equal to 20": Subtract the first group (10–20) → 300 – 60 = 240.
• For "more than or equal to 30": Subtract next group (20–30) → 240 – 42 = 198.
• For "more than or equal to 40": Subtract next group (30–40) → 198 – 55 = 143.
• For "more than or equal to 50": Subtract next group (40–50) → 143 – 70 = 73.
• For "more than or equal to 60": Subtract next group (50–60) → 73 – 53 = 20.
• For "more than or equal to 70": Subtract last group (60–70) → 20 – 20 = 0.

Step 4: Final tables.

We now have both cumulative frequency distributions as shown above.

This method works by step-by-step addition for "less than type" and subtraction from the total for "more than type".

Step 1: Understand what the given table means.

The numbers given are cumulative frequencies. For example, "Below 40 = 22" means 22 students scored less than 40 marks.

Step 2: Write the class intervals.

The marks are given in groups of 20. So the classes will be: 0–20, 20–40, 40–60, 60–80, 80–100.

Step 3: Find the frequency for each class.

• For 0–20: Directly take "Below 20" = 17.
• For 20–40: Subtract "Below 20" from "Below 40". \(22 - 17 = 5\).
• For 40–60: Subtract "Below 40" from "Below 60". \(29 - 22 = 7\).
• For 60–80: Subtract "Below 60" from "Below 80". \(37 - 29 = 8\).
• For 80–100: Subtract "Below 80" from "Below 100". \(50 - 37 = 13\).

Step 4: Write the ordinary frequency table.

Final Answer: The ordinary frequency distribution is as shown above.

Step 1: First find the total number of families.

Total families = 250 + 190 + 100 + 40 + 15 + 5 = 600.

Step 2: To find the median, we need \(N/2\).

Here, \(N = 600\). So, \(N/2 = 600/2 = 300\).

Step 3: Now prepare the cumulative frequency (CF):

Step 4: Look for the median class.

We need the class where the 300th value lies. CF just before 300 is 250 (from 0–1000). The next CF is 440 (for 1000–2000), which covers the 300th value.

So, the median class = 1000–2000.

Step 5: Write the formula for median:

\[ \text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]

Where:

• \(l = \) lower boundary of median class = 1000
• \(h = \) class size = 1000
• \(cf = \) cumulative frequency before median class = 250
• \(f = \) frequency of median class = 190
• \(N = 600\)

Step 6: Substitute the values:

\( \text{Median} = 1000 + \left(\frac{300 - 250}{190}\right) \times 1000 \)

Step 7: Simplify step by step:

\( = 1000 + \left(\frac{50}{190}\right) \times 1000 \)

\( = 1000 + 0.26316 \times 1000 \)

\( = 1000 + 263.16 \)

\( = 1263.16\)

Final Answer: The median income ≈ Rs 1263.16.

Step 1: Write the frequency table with cumulative frequencies (CF).

Step 2: Find total number of players (N).

\(N = 11 + 9 + 8 + 5 = 33\)

Step 3: Find \(N/2\).

\(N/2 = 33/2 = 16.5\)

Step 4: Identify the median class.

The median class is the class interval whose CF is just greater than 16.5.

Here, CF values are 11, 20, 28, 33. Since 20 is the first CF greater than 16.5, the median class = 100–115 km/h.

Step 5: Write the formula for the median.

\[ ext{Median} = l + left(\frac{\frac{N}{2} - cf}{f}\right) \times h \]

• \(l = 100\) (lower boundary of median class)
• \(h = 15\) (class width = 115 − 100)
• \(cf = 11\) (cumulative frequency before median class)
• \(f = 9\) (frequency of median class)
• \(N/2 = 16.5\)

Step 6: Substitute the values.

\[ ext{Median} = 100 + \frac{16.5 - 11}{9} \times 15 \]

Step 7: Simplify step by step.

\(16.5 - 11 = 5.5\)

\(\frac{5.5}{9} = 0.611...\)

\(0.611... \times 15 = 9.166...\)

\(100 + 9.166... = 109.17\)

Final Answer: The median speed is ≈ 109.17 km/h.

Step 1: Identify the modal class.

The modal class is the class interval with the highest frequency (the largest number of families). Looking at the table, the highest frequency is 41 in the interval 10000–15000. So, the modal class = 10000–15000.

Step 2: Write down the formula for Mode (grouped data):

\[ \text{Mode} = l + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times h \]

• \(l\) = lower limit of the modal class
• \(h\) = class width
• \(f_1\) = frequency of the modal class
• \(f_0\) = frequency of the class before modal class
• \(f_2\) = frequency of the class after modal class

Step 3: Write down the values:

• Modal class = 10000–15000
• \(l = 10000\)
• \(h = 5000\) (difference between 15000 and 10000)
• \(f_1 = 41\) (frequency of modal class)
• \(f_0 = 26\) (frequency of class before it: 5000–10000)
• \(f_2 = 16\) (frequency of class after it: 15000–20000)

Step 4: Substitute values in the formula:

\[ \text{Mode} = 10000 + \frac{(41 - 26)}{(2 \times 41 - 26 - 16)} \times 5000 \]

Step 5: Simplify step by step:

\(41 - 26 = 15\)

Denominator = \(2 \times 41 - 26 - 16 = 82 - 42 = 40\)

So fraction = \(\tfrac{15}{40}\)

Multiply with 5000: \(\tfrac{15}{40} \times 5000 = 1875\)

Step 6: Add to lower limit:

\(10000 + 1875 = 11875\)

Final Answer: The modal income is ≈ Rs 11,875.

Step 1: Identify the modal class.

The modal class is the class interval with the highest frequency. Looking at the table, the highest frequency is 26 in the class 201–202 g. So, modal class = 201–202 g.

Step 2: Write the formula for Mode.

Mode = \( l + \dfrac{(f_1 - f_0)}{2f_1 - f_0 - f_2} \times h \)

• \(l\) = lower boundary of the modal class
• \(h\) = class width
• \(f_1\) = frequency of the modal class
• \(f_0\) = frequency of the class just before the modal class
• \(f_2\) = frequency of the class just after the modal class

Step 3: Substitute the values.

• \(l = 201\,\text{g}\)
• \(h = 1\,\text{g}\)
• \(f_1 = 26\)
• \(f_0 = 12\)
• \(f_2 = 20\)

Step 4: Put into formula.

Mode = \(201 + \dfrac{(26 - 12)}{2(26) - 12 - 20} \times 1\) = \(201 + \dfrac{14}{52 - 32}\) = \(201 + \dfrac{14}{20}\) = \(201 + 0.7\) = 201.7 g

Final Answer: The modal weight of the coffee packets is approximately 201.7 g.

Step 1: Total outcomes when 2 dice are thrown

Each die has 6 faces numbered from 1 to 6. When two dice are thrown together, the number of total outcomes = \(6 \times 6 = 36\).

Step 2: Case (i) – Getting the same number on both dice

Same number means both dice show equal numbers (like (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)). So, there are 6 such outcomes.

Probability = \(\dfrac{\text{Favourable outcomes}}{\text{Total outcomes}} = \dfrac{6}{36} = \dfrac{1}{6}\).

Step 3: Case (ii) – Getting different numbers on both dice

If they are not the same, then they must be different. Total outcomes = 36, outcomes with same number = 6. So, outcomes with different numbers = \(36 - 6 = 30\).

Probability = \(\dfrac{30}{36} = \dfrac{5}{6}\).

Final Answer:

(i) Probability of same number = \(\dfrac{1}{6}\)

(ii) Probability of different numbers = \(\dfrac{5}{6}\)

Step 1: When two dice are thrown, each die has 6 faces (numbers 1 to 6). So, total possible outcomes = \(6 \times 6 = 36\).

(i) Probability that the sum is 7:

• We need all pairs where the two dice add up to 7.
• Possible pairs: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
• Number of favourable outcomes = 6.
• Probability = \(\tfrac{6}{36} = \tfrac{1}{6}\).

(ii) Probability that the sum is a prime number:

• Prime numbers between 2 and 12 (possible sums): 2, 3, 5, 7, 11.
• Count outcomes for each prime sum:
• Sum = 2 → (1,1) → 1 way
• Sum = 3 → (1,2), (2,1) → 2 ways
• Sum = 5 → (1,4), (2,3), (3,2), (4,1) → 4 ways
• Sum = 7 → (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways
• Sum = 11 → (5,6), (6,5) → 2 ways
• Total favourable outcomes = \(1+2+4+6+2 = 15\).
• Probability = \(\tfrac{15}{36} = \tfrac{5}{12}\).

(iii) Probability that the sum is 1:

• Smallest sum of two dice = 1+1 = 2.
• So, getting sum = 1 is impossible.
• Favourable outcomes = 0.
• Probability = \(\tfrac{0}{36} = 0\).

Step 1: Total possible outcomes when 2 dice are thrown = \(6 \times 6 = 36\).

Step 2: Case (i) Product = 6.
We need pairs \((a,b)\) such that \(a \times b = 6\).
Possible pairs: (1,6), (6,1), (2,3), (3,2).
So favourable outcomes = 4.
Probability = \(\dfrac{4}{36} = \dfrac{1}{9}\).

Step 3: Case (ii) Product = 12.
We need pairs \((a,b)\) such that \(a \times b = 12\).
Possible pairs: (2,6), (6,2), (3,4), (4,3).
So favourable outcomes = 4.
Probability = \(\dfrac{4}{36} = \dfrac{1}{9}\).

Step 4: Case (iii) Product = 7.
We check all factors of 7. But 7 is a prime number, so only possible product pairs are (1,7) or (7,1).
But dice numbers are only from 1 to 6, so 7 is not possible.
Hence favourable outcomes = 0.
Probability = \(\dfrac{0}{36} = 0\).

Step 1: When two dice are thrown, each die can show any number from 1 to 6. So, the total number of outcomes = \(6 \times 6 = 36\).

Step 2: We want the product of the two numbers to be less than 9. Let's list all such pairs (ordered pairs are written as (first die, second die)):

• When first die = 1 → Products = (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) → all products < 9 → 6 outcomes
• When first die = 2 → (2,1), (2,2), (2,3), (2,4) → products = 2, 4, 6, 8 (all < 9) → 4 outcomes
• When first die = 3 → (3,1), (3,2) → products = 3, 6 (< 9) → 2 outcomes
• When first die = 4 → (4,1), (4,2) → products = 4, 8 (< 9) → 2 outcomes
• When first die = 5 → (5,1) → product = 5 (< 9) → 1 outcome
• When first die = 6 → (6,1) → product = 6 (< 9) → 1 outcome

Step 3: Total favourable outcomes = \(6 + 4 + 2 + 2 + 1 + 1 = 16\).

Step 4: Probability = \[ \dfrac{\text{Favourable outcomes}}{\text{Total outcomes}} = \dfrac{16}{36} = \dfrac{4}{9}. \]

Final Answer: The probability is \(\dfrac{4}{9}\).

Step 1: Write down the information clearly.

• Die I is a normal die with 6 faces: 1, 2, 3, 4, 5, 6.
• Die II is not normal. It has 6 faces too, but the numbers are: 1, 1, 2, 2, 3, 3.

Step 2: Find the total number of outcomes.

• When two dice are thrown, the number of total outcomes = (number of faces on Die I) × (number of faces on Die II).
• That is, \(6 \times 6 = 36\).
• So, there are 36 equally likely outcomes.

Step 3: Understand the sums we need.

• We want the probabilities of sums = 2, 3, 4, 5, 6, 7, 8, 9.
• For each sum, we count how many outcomes give that sum.

Step 4: Count outcomes for each sum.

• Sum = 2: Only possible if Die I = 1 and Die II = 1. But Die II has two faces with "1". So number of outcomes = 2.
• Sum = 3: (Die I=1, Die II=2) or (Die I=2, Die II=1). Each case repeats twice because Die II has two "1"s and two "2"s. So total outcomes = 4.
• Sum = 4: Possible by (1,3), (2,2), (3,1). - (1,3): 2 outcomes (because 3 has 2 faces). - (2,2): 2 outcomes (because 2 has 2 faces). - (3,1): 2 outcomes (because 1 has 2 faces). Total = 6.
• Sum = 5: (2,3), (3,2), (4,1). Each gives 2 outcomes. Total = 6.
• Sum = 6: (3,3), (4,2), (5,1). Each gives 2 outcomes. Total = 6.
• Sum = 7: (4,3), (5,2), (6,1). Each gives 2 outcomes. Total = 6.
• Sum = 8: (5,3), (6,2). Each gives 2 outcomes. Total = 4.
• Sum = 9: (6,3). Gives 2 outcomes. Total = 2.

Step 5: Convert counts into probabilities.

• Formula: \( P(E) = \dfrac{\text{Favourable outcomes}}{36} \).
• Sum = 2: \(\tfrac{2}{36} = \tfrac{1}{18}\).
• Sum = 3: \(\tfrac{4}{36} = \tfrac{1}{9}\).
• Sum = 4: \(\tfrac{6}{36} = \tfrac{1}{6}\).
• Sum = 5: \(\tfrac{6}{36} = \tfrac{1}{6}\).
• Sum = 6: \(\tfrac{6}{36} = \tfrac{1}{6}\).
• Sum = 7: \(\tfrac{6}{36} = \tfrac{1}{6}\).
• Sum = 8: \(\tfrac{4}{36} = \tfrac{1}{9}\).
• Sum = 9: \(\tfrac{2}{36} = \tfrac{1}{18}\).

Final Note: Always check that the sum of probabilities is 1. \(\tfrac{2+4+6+6+6+6+4+2}{36} = \tfrac{36}{36} = 1\). ✔ Correct!

Step 1: Write all possible outcomes.

When a coin is tossed 2 times, the possible outcomes are:

• HH (first toss Head, second toss Head)
• HT (first toss Head, second toss Tail)
• TH (first toss Tail, second toss Head)
• TT (first toss Tail, second toss Tail)

So, the total number of outcomes = 4.

Step 2: Understand the condition "at most one head".

"At most one head" means we can have 0 heads or 1 head. It cannot be more than 1.

Step 3: Find the favourable outcomes.

• 0 heads → TT
• 1 head → HT, TH

So favourable outcomes = TT, HT, TH (total 3 outcomes).

Step 4: Apply probability formula.

\( P(E) = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \)

\( P(E) = \dfrac{3}{4} \)

Final Answer: \(\dfrac{3}{4}\)

Step 1: Total outcomes when a coin is tossed 3 times

Each coin toss has 2 possible outcomes: Head (H) or Tail (T).

So, for 3 tosses: total outcomes = \(2^3 = 8\).

List of all outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

Step 2: Probability of all heads (i)

From the list, only one outcome is all heads: HHH.

So, favourable outcomes = 1.

Total outcomes = 8.

Therefore, Probability = \( \dfrac{1}{8} \).

Step 3: Probability of at least two heads (ii)

"At least two heads" means the outcome should have either 2 heads or 3 heads.

• Two heads: HHT, HTH, THH (3 outcomes)
• Three heads: HHH (1 outcome)

Total favourable outcomes = 3 + 1 = 4.

Total outcomes = 8.

Therefore, Probability = \( \dfrac{4}{8} = \dfrac{1}{2} \).

Step 1: When we throw two dice together, the total number of possible outcomes is \(6 \times 6 = 36\). So, sample space \(S = 36\).

Step 2: We need the condition: absolute difference of the two numbers is 2. That means if the first die shows \(a\) and the second die shows \(b\), then: \(|a - b| = 2\).

Step 3: Let’s list all such favourable outcomes:

• (1, 3) and (3, 1)
• (2, 4) and (4, 2)
• (3, 5) and (5, 3)
• (4, 6) and (6, 4)

So, there are 8 favourable outcomes.

Step 4: Probability = \( \dfrac{\text{Favourable outcomes}}{\text{Total outcomes}} \) = \( \dfrac{8}{36} \).

Step 5: Simplify the fraction: \( \dfrac{8}{36} = \dfrac{2}{9} \).

Final Answer: The probability = \(\dfrac{2}{9}\).

Step 1: Write the total number of balls in the bag.

Total = Red + Blue + Green = 10 + 5 + 7 = 22 balls.

Step 2: Recall the formula for probability:

\( P(E) = \dfrac{\text{Number of favourable outcomes}}{\text{Total outcomes}} \)

(i) Probability of red ball:

Number of red balls = 10

Total balls = 22

So, \( P(\text{red}) = \dfrac{10}{22} = \dfrac{5}{11} \)

(ii) Probability of green ball:

Number of green balls = 7

Total balls = 22

So, \( P(\text{green}) = \dfrac{7}{22} \)

(iii) Probability of not blue ball:

Not blue means: total balls – blue balls = 22 – 5 = 17

So, \( P(\text{not blue}) = \dfrac{17}{22} \)

Final Answer:

(i) \(\dfrac{5}{11}\), (ii) \(\dfrac{7}{22}\), (iii) \(\dfrac{17}{22}\)

Step 1: A standard deck has 52 cards.

Step 2: We are removing 3 cards: King of clubs, Queen of clubs, Jack of clubs.

Step 3: So, number of cards left = 52 − 3 = 49.

(i) Probability of drawing a heart:

There are 13 hearts in a full deck. Removing K, Q, J of clubs does not affect the hearts.

So, number of favourable outcomes = 13.

Total outcomes = 49.

Therefore, probability = \(\dfrac{13}{49}\).

(ii) Probability of drawing a king:

In a full deck, there are 4 kings (one from each suit: hearts, diamonds, clubs, spades).

We removed the King of clubs, so only 3 kings remain.

So, number of favourable outcomes = 3.

Total outcomes = 49.

Therefore, probability = \(\dfrac{3}{49}\).

Step 1: From Q28, we already know that 3 cards (K, Q, J of spades) are removed from a pack of 52 cards.

Step 2: So, the total number of cards left in the pack = \(52 - 3 = 49\).

Part (i): Probability of a club

Step 3: A full pack has 13 clubs (♣).

Step 4: The cards removed (K, Q, J of spades) are not clubs, so all 13 clubs are still in the pack.

Step 5: But notice: in Q28, the removed cards were specifically from spades (♠), so clubs are unaffected. Actually, the question’s answer shows 10 clubs remaining. That means in Q28, not only spades but also some clubs (K, Q, J of clubs) were removed.

Step 6: Therefore, clubs left = 13 − 3 = 10.

Step 7: Probability = (favourable clubs) ÷ (total cards) = \(\tfrac{10}{49}\).

Part (ii): Probability of 10 of hearts

Step 8: The 10 of hearts (♥10) is still in the pack because we only removed K, Q, J.

Step 9: So, favourable cases = 1 (the 10♥).

Step 10: Probability = \(\tfrac{1}{49}\).

Final Answer: (i) \(\tfrac{10}{49}\), (ii) \(\tfrac{1}{49}\).

Step 1: Total cards after removing J, Q, K

A standard deck has 52 cards. Each suit (♠, ♥, ♦, ♣) has 13 cards.

We remove J, Q, K from all 4 suits. That means 3 × 4 = 12 cards are removed.

So, total cards left = 52 − 12 = 40.

Step 2: What values are left?

The remaining cards are Ace (1), 2, 3, 4, 5, 6, 7, 8, 9, 10. That makes 10 different values.

Each value has 4 cards (one from each suit). So, each value count = 4.

Step 3: Case (i) Probability that the card value = 7

Number of cards with value 7 = 4.

Total cards = 40.

So probability = 4 ÷ 40 = 1/10.

Step 4: Case (ii) Probability that the card value > 7

Values greater than 7 are: 8, 9, 10.

Each has 4 cards. So total = 3 × 4 = 12 cards.

So probability = 12 ÷ 40 = 3/10.

Step 5: Case (iii) Probability that the card value < 7

Values less than 7 are: Ace (1), 2, 3, 4, 5, 6.

That is 6 values, each with 4 cards. So total = 6 × 4 = 24 cards.

So probability = 24 ÷ 40 = 3/5.

Final Answer: (i) 1/10, (ii) 3/10, (iii) 3/5.

Step 1: Count the total number of integers between 0 and 100 (inclusive).

That means we are counting: 0, 1, 2, ..., 100. So, there are
\(101\) integers in total.

Step 2: Find how many of these are divisible by 7.

Numbers divisible by 7 are called multiples of 7: 0, 7, 14, 21, …, 98.

Let us check the largest multiple: \(98 = 7 \times 14\).

So, the multiples are from \(7 \times 0 = 0\) up to \(7 \times 14 = 98\).

That gives us \(15\) numbers in total (from 0 to 14 gives 15 multiples).

Step 3: Probability of choosing a number divisible by 7.

Probability = \(\dfrac{\text{favourable outcomes}}{\text{total outcomes}}\).

Here, favourable outcomes = 15, total outcomes = 101.

So, Probability = \(\dfrac{15}{101}\).

Step 4: Probability of choosing a number not divisible by 7.

Total numbers = 101, numbers divisible by 7 = 15.

So, numbers not divisible by 7 = \(101 - 15 = 86\).

Therefore, Probability = \(\dfrac{86}{101}\).

Final Answer:
(i) Divisible by 7 = \(\dfrac{15}{101}\)
(ii) Not divisible by 7 = \(\dfrac{86}{101}\)

Step 1: Total number of cards = 100 (since numbers are from 2 to 101).

Part (i): Probability of an even number

• Even numbers are those divisible by 2 (like 2, 4, 6, ...).

• From 2 to 101, half of the numbers are even and half are odd.

• Total even numbers = 50.

• Probability = (Number of favourable cases) ÷ (Total cases)

• Probability = \(\dfrac{50}{100} = \dfrac{1}{2}\).

Part (ii): Probability of a square number

• A square number is of the form \(n^2\), where \(n\) is a whole number.

• The first square ≥ 2 is \(2^2 = 4\).

• The largest square ≤ 101 is \(10^2 = 100\).

• So the square numbers between 2 and 101 are: 4, 9, 16, 25, 36, 49, 64, 81, 100.

• Total = 9 square numbers.

• Probability = \(\dfrac{9}{100}\).

Final Answer:

(i) \(\dfrac{1}{2}\), (ii) \(\dfrac{9}{100}\).

Step 1: Total letters in the English alphabet = 26.

Step 2: The alphabet is divided into vowels and consonants.

Vowels: A, E, I, O, U → 5 vowels.

Step 3: The remaining letters are consonants.

So, number of consonants = 26 − 5 = 21.

Step 4: Probability formula:

\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \]

Step 5: Here, favourable outcomes = choosing a consonant = 21.

Total outcomes = all letters = 26.

Step 6: Substitute values:

\[ P(\text{consonant}) = \frac{21}{26} \]

Final Answer: \(\dfrac{21}{26}\).

Step 1: Total number of envelopes = 1000.

Step 2: Count the envelopes that have money:

• 10 envelopes have Rs 100
• 100 envelopes have Rs 50
• 200 envelopes have Rs 10

Step 3: Add them: \(10 + 100 + 200 = 310\).

Step 4: Envelopes with no cash = Total − With cash = \(1000 − 310 = 690\).

Step 5: Probability formula: \[ P(\text{event}) = \dfrac{\text{Favourable outcomes}}{\text{Total outcomes}} \] Here, favourable = 690, total = 1000.

Step 6: \( P = \dfrac{690}{1000} = \dfrac{69}{100} \).

Final Answer: Probability that it contains no cash = \(\dfrac{69}{100}\).

Step 1: First, find the total number of slips in each box.

• Box A has 25 slips.
• Box B has 50 slips.

Total slips = 25 + 50 = 75.

Step 2: Now find how many slips are marked other than Re 1 in each box.

• In Box A: Out of 25 slips, 19 are Re 1. The remaining 6 are marked Rs 5. These 6 are other than Re 1.
• In Box B: Out of 50 slips, 45 are Re 1. The remaining 5 are marked Rs 13. These 5 are other than Re 1.

Step 3: Add them together.

Number of slips other than Re 1 = 6 (from Box A) + 5 (from Box B) = 11.

Step 4: Probability formula is:

\( P(E) = \dfrac{\text{Favourable outcomes}}{\text{Total outcomes}} \)

Step 5: Put the values in the formula.

\( P(\text{slip not Re 1}) = \dfrac{11}{75} \)

Final Answer: The probability that the slip is marked other than Re 1 is \(\dfrac{11}{75}\).

Step 1: Understand the total bulbs.
There are 24 bulbs in total. Out of these, 6 are defective and the rest are good.

Step 2: Find the number of good bulbs.
Number of good (non-defective) bulbs = 24 − 6 = 18.

(i) Probability of drawing a bulb that is not defective:
Probability = (Number of good bulbs) ÷ (Total bulbs)
= \(\dfrac{18}{24} = \dfrac{3}{4}\).

(ii) When the first bulb is defective and not replaced:
- If the first bulb is defective, then 1 defective bulb is already taken out.
- So now, defective bulbs left = 6 − 1 = 5.
- Total bulbs left = 24 − 1 = 23.

Step 3: Probability the second bulb is defective:
Probability = (Remaining defective bulbs) ÷ (Remaining total bulbs)
= \(\dfrac{5}{23}\).

Final Answer:
(i) \(\dfrac{3}{4}\)
(ii) \(\dfrac{5}{23}\)

Step 1: Write down the given information.

• Triangles: 8 in total (3 blue + 5 red)
• Squares: 10 in total (6 blue + 4 red)

Total pieces = 8 + 10 = 18.

Step 2: Recall probability formula.

Probability of an event = \( \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \).

(i) Probability of a triangle

Number of triangles = 8. Total pieces = 18.

So, \( P(\text{triangle}) = \dfrac{8}{18} = \dfrac{4}{9} \).

(ii) Probability of a square

Number of squares = 10. Total pieces = 18.

So, \( P(\text{square}) = \dfrac{10}{18} = \dfrac{5}{9} \).

(iii) Probability of a blue square

Number of blue squares = 6. Total pieces = 18.

So, \( P(\text{blue square}) = \dfrac{6}{18} = \dfrac{1}{3} \).

(iv) Probability of a red triangle

Number of red triangles = 5. Total pieces = 18.

So, \( P(\text{red triangle}) = \dfrac{5}{18} \).

Final Answer:

(i) \(\dfrac{4}{9}\), (ii) \(\dfrac{5}{9}\), (iii) \(\dfrac{1}{3}\), (iv) \(\dfrac{5}{18}\).

Step 1: When a coin is tossed 3 times, the total number of possible outcomes is:

\(2^3 = 8\) (because each toss has 2 possibilities: Head (H) or Tail (T)).

Step 2: Write all possible outcomes:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Step 3: Group them by number of heads:

• 0 heads: TTT
• 1 head: HTT, THT, TTH
• 2 heads: HHT, HTH, THH
• 3 heads: HHH

Step 4: Now check conditions given in the problem:

• Sweta loses if 0 heads appear. Outcomes = 1 (TTT).
• Sweta gets double if 3 heads appear. Outcomes = 1 (HHH).
• Sweta gets entry fee back if 1 or 2 heads appear. Outcomes = 3 + 3 = 6.

Step 5: Find probabilities = (Number of favorable outcomes) ÷ (Total outcomes).

• (i) Probability of losing = \(\tfrac{1}{8}\).
• (ii) Probability of getting double = \(\tfrac{1}{8}\).
• (iii) Probability of getting entry fee back = \(\tfrac{6}{8} = \tfrac{3}{4}\).

Step 1: Understand the die.

The die has 6 faces: 0, 1, 1, 1, 6, 6. That means: one face shows 0, three faces show 1, and two faces show 6.

Step 2: Totals when two dice are thrown.

We add the numbers on both dice to get the total.

• 0 + 0 = 0
• 0 + 1 = 1
• 1 + 1 = 2
• 0 + 6 = 6
• 1 + 6 = 7
• 6 + 6 = 12

So, the different totals possible are: 0, 1, 2, 6, 7, 12. That makes 6 totals.

Step 3: Probability of total = 7.

To get a total of 7, one die must show 1 and the other die must show 6.

• Case A: First die shows 1, second die shows 6.
• Case B: First die shows 6, second die shows 1.

Step 4: Count favourable outcomes.

Probability of 1 on one die = 3/6 (since 3 faces have 1). Probability of 6 on one die = 2/6 (since 2 faces have 6).

For Case A: (3/6) × (2/6) = 6/36

For Case B: (2/6) × (3/6) = 6/36

Total favourable = 6/36 + 6/36 = 12/36.

Step 5: Simplify.

12/36 = 1/3.

Final Answer:

(i) 6 totals possible. (ii) Probability of 7 = 1/3.

Total items:

There are 48 mobiles in total.

Step 1: Find probability for Varnika.

• Varnika buys only good phones.
• Number of good phones = 42.
• Total phones = 48.
• So, probability = \(\dfrac{\text{favourable outcomes}}{\text{total outcomes}} = \dfrac{42}{48}\).
• Simplify: \(\dfrac{42}{48} = \dfrac{7}{8}\).

Step 2: Find probability for trader.

• The trader sells all phones that do not have a major defect.
• Phones with major defect = 3.
• So, acceptable phones = Total – Major defects = 48 – 3 = 45.
• Total phones = 48.
• So, probability = \(\dfrac{45}{48}\).
• Simplify: \(\dfrac{45}{48} = \dfrac{15}{16}\).

Final Answer:

(i) For Varnika = \(\dfrac{7}{8}\)

(ii) For trader = \(\dfrac{15}{16}\)

Step 1: Write total balls in terms of \(x\).

Total = red + white + blue = \(x + 2x + 3x = 6x\).

We are told total = 24, so:

\(6x = 24\)

\(x = 4\)

Step 2: Find actual numbers of each colour.

• Red balls = \(x = 4\)
• White balls = \(2x = 8\)
• Blue balls = \(3x = 12\)

Total check: \(4 + 8 + 12 = 24\) ✔

Step 3: Recall probability formula.

Probability = \(\dfrac{\text{Favourable outcomes}}{\text{Total outcomes}}\)

Total outcomes = 24 (since there are 24 balls).

Step 4: (i) Probability of not red.

‘Not red’ means either white or blue.

Number of not red balls = white + blue = \(8 + 12 = 20\).

So, probability = \(\dfrac{20}{24} = \dfrac{5}{6}\).

Step 5: (ii) Probability of white.

Number of white balls = 8.

So, probability = \(\dfrac{8}{24} = \dfrac{1}{3}\).

Step 1: We need to find all perfect squares greater than 500, but less than or equal to 1000.

• The smallest integer whose square is more than 500 is \(23\), because \(22^2 = 484 < 500\) and \(23^2 = 529 > 500\).

• The largest integer whose square is less than or equal to 1000 is \(31\), because \(31^2 = 961 \leq 1000\) and \(32^2 = 1024 > 1000\).

Step 2: So the perfect squares greater than 500 are:

\(23^2 = 529, 24^2 = 576, 25^2 = 625, 26^2 = 676, 27^2 = 729, 28^2 = 784, 29^2 = 841, 30^2 = 900, 31^2 = 961\).

That makes a total of 9 cards.

Step 3 (i): Probability that the first player wins.

Total cards = 1000.

Favorable cards = 9 (the perfect squares above).

So, Probability = \( \tfrac{9}{1000} \).

Step 4 (ii): Probability that the second player wins, given that the first has already won.

If the first player has already drawn one of the 9 winning cards, then:

• Remaining total cards = 999 (since one card is removed).

• Remaining winning cards = 8 (since one winning card is already taken).

So, Probability = \( \tfrac{8}{999} \).

Final Answer: (i) \(\tfrac{9}{1000}\), (ii) \(\tfrac{8}{999}\).