Question

If A and B are two events associated with a random experiment, then
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
If A, B, C are three events associated with a random experiment, then
$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)$

Two cards are drawn from a pack of 52 cards. What is the probability that either both are red or both are kings?

Answer 1

We have, $n\left(S\right)={}^{52}{C}_2=\frac{52\times 51}{2}=1326.$

Let A = event of getting both black cards and B = event of getting both queens

$A\cap B=$ event of getting queen of black cards

$n\left(A\right)={}^{26}{C}_2=\frac{26\times 25}{2\times 1}=325,n\left(B\right)={}^4{C}_2=\frac{4\times 3}{2\times 1}=6$ and

$n(A\cap B)={}^2{C}_2=1$

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{325}{1326}$

$P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{6}{1326}$

$P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{1}{1326}$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)=\frac{(325+6-1)}{1326}=\frac{330}{1326}=\frac{55}{221}$