A fair coin is tossed twice. Let A be the event that at least one tail appears and B be the event that both head and tail appear. Then P[A/B], the probability of A given B, isa)1/4b)1/2c)2/3d)1Correct answer is option 'D'. Can you explain this answer?

Question

Answer

Understanding the Events
When a fair coin is tossed twice, the sample space (all possible outcomes) is:
- HH (two heads)
- HT (head and tail)
- TH (tail and head)
- TT (two tails)
Thus, the sample space contains 4 outcomes.
Defining Events A and B
- Event A: At least one tail appears. The outcomes that satisfy this are HT, TH, and TT.
- Event B: Both head and tail appear. The outcomes that satisfy this are HT and TH.
Calculating Probabilities
Let's identify the outcomes for events A and B:
- A = {HT, TH, TT}
- B = {HT, TH}
Now, we need to find P[A/B], which is the probability of A occurring given that B has occurred.
Finding P[A/B]
Using the definition of conditional probability:
P[A/B] = P(A and B) / P(B)
- P(A and B): The outcomes in both A and B are {HT, TH}. Therefore, P(A and B) = 2/4 = 1/2.
- P(B): The outcomes in B = {HT, TH}, so P(B) = 2/4 = 1/2.
Now, substituting these values into the formula:
P[A/B] = (1/2) / (1/2) = 1
Conclusion
Thus, the correct answer is option 'D', which states that P[A/B] = 1. This means that if both head and tail appear, it is certain that at least one tail appears.

Answer

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