Question

A box contains $30$ toys of same size in which $10$ toys are white and all the remaining toys are blue. A toy is drawn at random horn the box and it is replaced in the box after noting down its colour. If $5$ toys are drawn in tills way, then the probability of getting atmost $2$ white toys is

• A. $\left(\frac{6}{9}\right)^{2}$
• B. $\left(\frac{8}{9}\right)^{2}$
• C. $\left(\frac{7}{9}\right)^{2}$
• D. $\left(\frac{2}{3}\right)^{5}$

Answer 1

Answer: (B)

$\left(\frac{8}{9}\right)^{2}$

Answer 2

Required probability
$=$ No white ball $+$ one white ball + two white ball
$=\left(\frac{2}{3}\right)^{5}+{ }^{5} C_{1}\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)^{1}+{ }^{5} C_{2}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2}$
$=\left(\frac{1}{3}\right)^{5}\left[2^{5}+5 \cdot 2^{4}+10 \cdot 2^{3}\right]$
$=\left(\frac{1}{3}\right)^{5}[32+80+80]=\frac{192}{3^{5}}=\frac{64 \times 3}{3^{5}}$
$=\frac{64}{3^{4}}=\frac{8 \times 8}{3^{2} \times 3^{2}}=\left(\frac{8}{9}\right)^{2}$