-2
$\begingroup$

Question: Suppose a family has 2 children, one of which is a boy. What is the probability that both children are boys?

My attempt:

P(both are boys | one is a boy)

= P(both are boys AND one is a boy) / P(one is a boy)

= P(one is a boy | both are boys) * P(both are boys) / P(one is a boy)

= 1 * (1/4) / (1 - 2(1/2)^2)

= 1/2

But the real answer is 1/3. Why?

$\endgroup$

marked as duplicate by rlartiga, Ian Miller, Adam Hughes, TastyRomeo, Davide Giraudo Jan 3 '17 at 15:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Where do you get $1 - 2(1/2)^2$ in the numerator? $\endgroup$ – Mees de Vries Jan 3 '17 at 14:13
  • $\begingroup$ Please provide all given information (i.e. the probability of a child being a boy) $\endgroup$ – pseudoeuclidean Jan 3 '17 at 14:14
  • 1
    $\begingroup$ You mean the denominator? I assumed P(one is a boy) = one minus the probability of two boys or two girls = 1 - (1/2)^2 - (1/2)^2 = 1 - 2(1/2)^2 = 1 - 1/2 = 1/2 $\endgroup$ – user7368066 Jan 3 '17 at 14:15
  • 1
    $\begingroup$ @user7368066, yes, denominator, my bad. Note that you are using "one boy" in two different ways: in the entire computation, "one boy" means "at least one boy", but then to compute the denominator you interpret it to mean "precisely one boy". If you use the same interpretation consistently, you get the right answer. $\endgroup$ – Mees de Vries Jan 3 '17 at 14:16
0
$\begingroup$

Your second to last line should be

$$\frac{1\cdot\frac14}{1-\left(\frac12\right)^2}$$

So you wrote one $2$ too many. There is no reason for the $2$ in (1 - 2(1/2)^2)

$\endgroup$
  • $\begingroup$ Where is my error in: P(one is a boy) = one minus the probability of two boys or two girls = 1 - (1/2)^2 - (1/2)^2 = 1 - 2(1/2)^2 = 1 - 1/2 = 1/2 $\endgroup$ – user7368066 Jan 3 '17 at 14:16
  • 1
    $\begingroup$ @user7368066 One boy means "at least one boy" (otherwise, P(one is a boy | both are boys)=0) . So don't subtract the probability of two boys. $\endgroup$ – 5xum Jan 3 '17 at 14:17
0
$\begingroup$

Assuming boys and girls have same probability to be born. You have the following cases for the two children: BB, BG, GB (B is boy, G ir girl). From those 3 only 1 is two boys so the probability is $1/3$ as stated.

$\endgroup$
  • $\begingroup$ I don't see why marked down by people, it seems to be an answer. $\endgroup$ – Cato Jan 3 '17 at 14:17
  • $\begingroup$ Mind explaining why you donwvote? $\endgroup$ – Jesús Ros Jan 3 '17 at 14:17
  • $\begingroup$ You answer doesn't address why the OP's attempt is incorrect. You've just given a standard answer to a question which has been asked many times on this site. $\endgroup$ – Ian Miller Jan 3 '17 at 23:25

Not the answer you're looking for? Browse other questions tagged or ask your own question.