Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is :

Question

Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is : (A) (1/11) (B) (1/17) (C) (1/10) (D) (1/12)

Tardigrade Admin · Accepted Answer

P(B) = P(G) = 1/2 Required Proballity = ( textall 4 textgirls/( textall 4 textgirls)+( textexactly 3 textgirls+1 textboy)+ ( textexactly 2 textgirls+2 textboy)) = (((1/2))4/((1)2)4 + 4C3((1)2)4 +4C2 ((1/2))4 = (1/11)

Q. Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is :

$\frac{1}{11}$

$\frac{1}{17}$

$\frac{1}{10}$

$\frac{1}{12}$

$P (B) = P (G) = 1/2$
Required Proballity = $\frac{all 4 girls}{( all 4 girls ) + ( exactly 3 girls + 1 boy ) + ( exactly 2 girls + 2 boy )}$
$= \frac{( \frac{1}{2} ) ^{4}}{( \frac{1}{2} ) ^{4} + ^{4} C _{3} ( \frac{1}{2} ) ^{4} + ^{4} C _{2} ( \frac{1}{2} ) ^{4}} = \frac{1}{11}$

Q. Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is :

111​

171​

101​

121​

P(B)=P(G)=1/2 Required Proballity = (all4girls)+(exactly3girls+1boy)+(exactly2girls+2boy)all4girls​ =(21​)4+4C3​(21​)4+4C2​(21​)4(21​)4​=111​

$\frac{1}{11}$

$\frac{1}{17}$

$\frac{1}{10}$

$\frac{1}{12}$

$P (B) = P (G) = 1/2$
Required Proballity = $\frac{all 4 girls}{( all 4 girls ) + ( exactly 3 girls + 1 boy ) + ( exactly 2 girls + 2 boy )}$
$= \frac{( \frac{1}{2} ) ^{4}}{( \frac{1}{2} ) ^{4} + ^{4} C _{3} ( \frac{1}{2} ) ^{4} + ^{4} C _{2} ( \frac{1}{2} ) ^{4}} = \frac{1}{11}$