Probability — Class 12 Maths Solution

.: Let $P$ be the probability of drawn card is spade $= \cfrac{{13}}{{52}} = \cfrac{1}{4}$

$\therefore$ $q = 1 - \cfrac{1}{4} = \cfrac{3}{4}$

$X$ has a binomial distribution with $n = 5,p = \cfrac{1}{4},q = \cfrac{3}{4}$

$\Rightarrow P(X = r) = n{C_r}{q^{n - r}}{p^r}$
(i) $P$ (all five cards are spades) $= P(X = 5)$

${ = ^5}{C_5}{q^0}{p^5} = (1)(1){\left( {\cfrac{1}{4}} \right)^5} = {\left( {\cfrac{1}{4}} \right)^5} = \cfrac{1}{{1024}}$

(ii) $P$ (only 3 cards are spades) $= P(X = 3)$

${ = ^5}{C_3}{q^2}{p^3} = 10{\left( {\cfrac{3}{4}} \right)^2}{\left( {\cfrac{1}{4}} \right)^3} = 90 \times {\left( {\cfrac{1}{4}} \right)^5} = \cfrac{{90}}{{1024}} = \cfrac{{45}}{{512}}$

(iii) $P$ (none is spade) $= P(X = 0)$
${ = ^5}{C_0}{q^5}{p^0} = (1){\left( {\cfrac{3}{4}} \right)^5}(1) = {\left( {\cfrac{3}{4}} \right)^5} = \cfrac{{243}}{{1024}}$