A box contains 30 toys of the same size,in wh | vedclass

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(B) Total toys = $30$. White toys = $10$. Blue toys = $30 - 10 = 20$. Probability of drawing a white toy $(p)$ = $\frac{10}{30} = \frac{1}{3}$. Probab

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

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(B) Total toys = $30$. White toys = $10$. Blue toys = $30 - 10 = 20$. Probability of drawing a white toy $(p)$ = $\frac{10}{30} = \frac{1}{3}$. Probability of drawing a blue toy $(q)$ = $1 - p = 1 - \frac{1}{3} = \frac{2}{3}$. Since the toys are replaced,this follows a binomial distribution $B(n, p)$ with $n = 5$ and $p = \frac{1}{3}$. We need the probability of getting at most $2$ white toys,which is $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$. $P(X=k) = {}^nC_k p^k q^{n-k}$. $P(X=0) = {}^5C_0 (\frac{1}{3})^0 (\frac{2}{3})^5 = (\frac{2}{3})^5$. $P(X=1) = {}^5C_1 (\frac{1}{3})^1 (\frac{2}{3})^4 = 5 	imes \frac{1}{3} 	imes \frac{16}{81} = \frac{80}{243}$. $P(X=2) = {}^5C_2 (\frac{1}{3})^2 (\frac{2}{3})^3 = 10 	imes \frac{1}{9} 	imes \frac{8}{27} = \frac{80}{243}$. $P(X \le 2) = \frac{32}{243} + \frac{80}{243} + \frac{80}{243} = \frac{192}{243}$. Dividing numerator and denominator by $3$,we get $\frac{64}{81} = (\frac{8}{9})^2$.

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

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