I toss three coins together. The possible outcomes are no heads, 1 head, 2 heads and 3 heads. So, I say that probability of no heads is \(\dfrac{1}{4}\). What is wrong with this conclusion?
Step 1: When 3 coins are tossed together, each coin can show either Head (H) or Tail (T).
Step 2: Total number of possible outcomes = \(2^3 = 8\).
Step 3: Write the sample space (all outcomes): {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
Step 4: "No heads" means all coins must be Tails. That happens only in one case: TTT.
Step 5: Number of favourable outcomes = 1 (only TTT).
Step 6: Probability formula is:
\( P(E) = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \)
Step 7: Substituting values:
\( P(\text{no heads}) = \dfrac{1}{8} \).
Step 8: The wrong reasoning was treating the 4 cases (no head, 1 head, 2 heads, 3 heads) as equally likely. But they are not. Each event has a different number of outcomes.
Final Answer: Probability of no heads is \(\dfrac{1}{8}\), not \(\dfrac{1}{4}\).