Multiple choice

A coin is tossed three times in succession. If $E$ is the event that there are at least two heads and $F$ is the event in which first throw is a head, then $P(E/F)$ is equal to:

  1. $3/4$
  2. $3/8$
  3. $1/2$
  4. $1/8$
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A Correct answer
Explanation

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

AI explanation

The sample space for tossing a coin three times has 8 outcomes. Event F, where the first throw is a head, consists of HHH, HHT, HTH, and HTT, giving 4 outcomes. Out of these, the outcomes with at least two heads (event E) are HHH, HHT, and HTH, providing 3 favorable cases. Using the conditional probability formula P(E/F), we get 3 divided by 4. The result is 3/4.

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