Question

Can anyone solve Question 11 ??​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{\huge{\underline{ english}}}$

$\bold{\underline{Question}}$

A Family has 2 children . Find the Probability that both are boys, if it is know that

(i) at least one of the children is a boy

(ii) the elder child is boy

$\bold{\underline{Answer}}$ According to the question we know that in a family has two children

Now,

let, B = Boy and G = Girl

then

S = {gg}{bb}{gb}{bg}

(i) When 1 children is boy

B' = {bb}{gb}{bg}

so,

P(B) = 3/4

(ii)

when both children are boys

B = {bb}

so,

P(B') = 1/4

Also,

$B \cap \: B' \: = (b, \: b) \\ \\ p(B \cap B') = \frac{1}{4}$

Now,

$p( \frac{B }{\: B'} ) = \frac{p(B \cap \: B')}{p(B')} \\ \\ = \frac{ \frac{1}{4} }{ \frac{3}{4} } = \frac{1}{3} \\ \\ p( \frac{B }{\: B'} ) = \frac{1}{3}$

Answer 2

Answer​ According to the question we know that in a family has two children 

Now,

let, B = Boy and G = Girl 

then 

S = {gg}{bb}{gb}{bg}

(i) When 1 children is boy 

B' = {bb}{gb}{bg}

so,

P(B) = 3/4 

(ii)

when both children are boys

B = {bb}

so,

P(B') = 1/4 

Also,

\begin{lgathered}B \cap \: B' \: = (b, \: b) \\ \\ p(B \cap B') = \frac{1}{4}\end{lgathered}B∩B′=(b,b)p(B∩B′)=41​​ 

Now,

\begin{lgathered}p( \frac{B }{\: B'} ) = \frac{p(B \cap \: B')}{p(B')} \\ \\ = \frac{ \frac{1}{4} }{ \frac{3}{4} } = \frac{1}{3} \\ \\ p( \frac{B }{\: B'} ) = \frac{1}{3}\end{lgathered}p(B′B​)=p(B′)p(B∩B′)​=43​41​​=31​p(B′B​)=31​​