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A family has 8 girls and 4 boys. A total of 2 children must speak on behalf of the family at a local benefit. What is the probability that at least one boy will be chosen?

1 Answer

Jul 6, 2018

Let's look at the probability that no boy is chosen.

Explanation:

First selection:
$P (g i r l) = \frac{8}{12} = \frac{2}{3}$ $P(girl)=8/12=2/3$

Second selection:
$P (g i r l) = \frac{7}{11}$ $P(girl)=7/11$

Combined probability of first AND second being a girl
(Remember AND means MULTIPLY):

$P (2 g i r l s) = P (0 b o y s) = \frac{2}{3} \times \frac{7}{11} = \frac{14}{33}$ $P(2 girls)=P(0boys)=2/3xx7/11=14/33$

The probability of at least one boy is then the complement of this, or in other words:

$P (1 or 2 b o y s) = 1 - P (2 g i r l s) = 1 - \frac{14}{33} = \frac{19}{33} \approx 58 %$ $P(1or 2 boys)=1-P(2 girls)=1-14/33=19/33~~58%$

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