A student says that if you throw a die, it will show up 1 or not 1. Therefore, the probability of getting 1 and the probability of getting ‘not 1’ each is equal to \(\dfrac{1}{2}\). Is this correct? Give reasons.
Step 1: When we throw a die, the sample space (all possible outcomes) is:
\(S = \{1, 2, 3, 4, 5, 6\}\)
So there are a total of 6 equally likely outcomes.
Step 2: Probability of getting a 1:
Only one outcome (1) is favorable.
So, \(P(1) = \dfrac{1}{6}\).
Step 3: Probability of getting ‘not 1’:
Here the favorable outcomes are \(\{2, 3, 4, 5, 6\}\).
That means there are 5 favorable outcomes.
So, \(P(\text{not 1}) = \dfrac{5}{6}\).
Step 4: Compare the two probabilities:
\(P(1) = \dfrac{1}{6}\), but \(P(\text{not 1}) = \dfrac{5}{6}\).
These are not equal.
Final Step: The student’s claim that both probabilities are \(\dfrac{1}{2}\) is false. The correct probabilities are \(\dfrac{1}{6}\) and \(\dfrac{5}{6}\).