Question

An urn contains $4$ red and $7$ blue balls. Two balls are drawn at random with replacement. Find the probability of getting
$\left(i\right)$ $2$ red balls,
$\left(ii\right)$ $2$ blue balls,
$\left(iii\right)$ one red and one blue ball.

Answer 1

On each drawing, total balls $=11$

Red balls $=4$,

Black balls $=7$

(i) $P($two red balls$)$ $=\frac{4}{11}\times \frac{4}{11}=\frac{16}{121}$

(ii) $P($two black balls$)$$=\frac{7}{11}\times \frac{7}{11}=\frac{49}{121}$

(iii) red ball can be in first or second draw

$\Rightarrow$ $P($one red and one black ball$)$ = $2\times \left(\frac{4}{11}\right)\left(\frac{7}{11}\right)=\frac{56}{121}$