Question

Find the probability that when a hand of $7$ cards is drawn from a well shuffled deck of $52$ cards, it contains

$\left(i\right)$ All Kings

$\left(ii\right)$ $3$ Kings

$\left(iii\right)$ at least $3$ Kings.

Answer 1

$\left(i\right)$ Required probability
Given : $7$ cards are to be chosen out of $52$ cards.
The possible outcomes are $={}^{52}{C}_7$

$n\left(S\right)={}^{52}{C}_7=\frac{52!}{7!(52-7)!}$

$=\frac{52!}{7!45!}\{\because {}^n{C}_r=\frac{n!}{r!(n-r)!}\}$

Let $A$ be the event that all kings are selected.
There are only $4$ kings in a pack of $52$ cards.
So, $4$ kings to be chosen out of $4$ and $3$ other cards are to be chosen out of remaining $48$ cards.

Hence required possible outcomes are $n\left(4\right)={}^4{C}_4{\times }^{48}{C}_3$

$n\left(A\right)=\frac{4!}{4!0!}\times \frac{48!}{3!(48-3)!}=1\times \frac{48!}{3!45!}$

$=\frac{48!}{3!45!}$

Probability that all kings selected = $\frac{\text{favourable outcomes}}{\text{Total outcomes}}$

$\Rightarrow P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$

$\Rightarrow P\left(A\right)=\frac{48!}{3!45!}\times \frac{7!45!}{52!}$

$\Rightarrow P\left(A\right)=\frac{48!\times 7!}{3!\times 52!}$

$=\frac{48!\times 7!}{3!\times 52\times 51\times 50\times 49\times 48!}$

$\Rightarrow P\left(A\right)=\frac{7!}{3!\times 52\times 51\times 50\times 49}=\frac{1}{7735}$

Hence the required probability that all kings are selected is $\frac{1}{7735}$

$\left(ii\right)$ Required probability
Given : $7$ cards are to be chosen out of $52$ cards.
The possible outcomes are $={}^{52}{C}_7$

$n\left(S\right)={}^{52}{C}_7=\frac{52!}{7!(52-7)!}$

$=\frac{52!}{7!45!}\{\because {}^n{C}_r=\frac{n!}{r!(n-r)!}\}$

Let $B$ be the event that $3$ kings are selected
There are only $4$ kings in a pack of $52$ card.
So, $3$ kings to be chosen out of $4$ and $4$ other cards are to be chosen out of remaining $48$ cards.

Hence required possible outcomes are $n\left(B\right)={}^4{C}_3{\times }^{48}{C}_4$

$n\left(B\right)=\frac{4!}{3!(4-3)!}\times \frac{48!}{4!(48-4)!}=4\times \frac{48!}{4!44!}$

$P(3$ kings selected $)=$ $\frac{\text{favourable outcomes}}{\text{Total outcomes}}$

$P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}$

$P\left(B\right)=\frac{4\times \frac{48!}{4!44!}}{\frac{52!}{7!45!}}$

$P\left(B\right)=\frac{4\times 48!\times 7!\times 45!}{4!\times 44!\times 52!}$

$P\left(B\right)=\frac{4\times 48!\times 7!\times 45!}{4!\times 44!\times 52\times 51\times 50\times 49\times 48!}$

$P\left(B\right)=\frac{4\times 7!\times 45}{4!\times 52\times 51\times 50\times 49}$

$P\left(B\right)=\frac{4\times 7\times 6\times 5\times 4!\times 45}{52\times 51\times 50\times 49\times 4!}=\frac{9}{1547}$

Hence the required probability that all $3$ kings is selected $\frac{9}{1547}$

$\left(iii\right)$ Simplification of the given data
Given : $7$ cards are to be chosen out of $52$ cards.
The possible outcomes are $={}^{52}{C}_7$

$n\left(S\right)={}^{52}{C}_7=\frac{52!}{7!(52-7)!}$

$=\frac{52!}{7!45!}\{\because {}^n{C}_r=\frac{n!}{r!(n-r)!}\}$

Let $A$ be the event that all kings are selected
There are only $4$ kings in a pack of $52$ cards.
So, $4$ kings to be chosen out of $4$ and $3$ other cards are to be chosen out of remaining $48$ cards.

Hence required possible outcomes are $n\left(A\right){=}^4{C}_4{\times }^{48}{C}_3$

$n\left(A\right)=\frac{4!}{4!0!}\times \frac{48!}{3!(48-3)!}=1\times \frac{48!}{3!45!}$
$=\frac{48!}{3!45!}$

Probability that all kings selected $=\frac{\text{favourable outcomes}}{\text{Total outcomes}}$

$\Rightarrow P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$

$\Rightarrow P\left(A\right)=\frac{48!}{3!45!}\times \frac{7!45!}{52!}$

$\Rightarrow P\left(A\right)=\frac{48!\times 7!}{3!\times 52!}$

$=\frac{48!\times 7!}{3!\times 52\times 51\times 50\times 49\times 48!}$

$\Rightarrow P\left(A\right)=\frac{7!}{3!\times 52\times 51\times 50\times 49}=\frac{1}{7735}$ $...\left(i\right)$

Let $B$ be the event that $3$ kings are selected
There are only $4$ kings in a pack of $52$ cards.
So, $3$ kings to be chosen out of $4$ and $4$ other cards are to be chosen out of remaining $48$ cards.

Hence the required possible outcomes are
$n\left(B\right){=}^4{C}_3{\times }^{48}{C}_4$

$n\left(B\right)=\frac{4!}{3!(4-3)!}\times \frac{48!}{4!(48-4)!}=4\times \frac{48!}{4!44!}$

$P(3$ kings selected $)$ $=\frac{\text{favourable outcomes}}{\text{Total outcomes}}$

$P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}$

$P\left(B\right)=\frac{4\times \frac{48!}{4!44!}}{\frac{52!}{7!45!}}$

$P\left(B\right)=\frac{4\times 48!\times 7!\times 45!}{4!\times 44!\times 52!}$

$P\left(B\right)=\frac{4\times 48!\times 7!\times 45!}{4!\times 44!\times 52\times 51\times 50\times 49\times 48!}$

$P\left(B\right)=\frac{4\times 7!\times 45}{4!\times 52\times 51\times 50\times 49}$

$P\left(B\right)=\frac{4\times 7\times 6\times 5\times 4!\times 45}{52\times 51\times 50\times 49\times 4!}=\frac{9}{1547}$ $...\left(ii\right)$

Required probability

Let $k$ be the event that at least $3$ kings selected.

$P$ $($at least $3$ kings$)=$ $P(3$ kings $)$ + $P(4$ kings $)$ $....\left(iii\right)$

Putting values $\left(i\right)$ and $\left(ii\right)$ in $\left(iii\right)$, we get

$P$ $($at least $3$ kings$)=$$\frac{9}{1547}+\frac{1}{7735}$

$P\left(k\right)=\frac{46}{7735}$