Multiple choice

A coin is tossed $3$ times. If $E$ denotes the event in which heads appear at least twice and $F$ denotes the event in which head comes first, then $P(\displaystyle {E}|{F})=$

  1. $\displaystyle \frac{3}{4}$
  2. $\displaystyle \frac{3}{8}$
  3. $\displaystyle \frac{1}{2}$
  4. $\displaystyle \frac{1}{8}$
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A Correct answer
Explanation

Sample space: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. F (head first): HHH, HHT, HTH, HTT. E (at least 2 heads): HHH, HHT, HTH, THH. E intersection F: HHH, HHT, HTH. P(E|F) = 3/4.

AI explanation

Using the conditional probability formula P(E|F) = P(E intersection F) / P(F), we first find P(F), the probability of getting a head on the first toss, which is 1/2. The event E intersection F represents getting at least two heads total and a head on the first toss, meaning the last two tosses must yield at least one head; the probability of this is 3/4 multiplied by 1/2, which is 3/8. Dividing 3/8 by 1/2 gives the final probability of 3/4.

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