Question

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

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

Two cards are drawn from a pack of $52$ cards, then the probability that either both are red or both are kings is

• A. $\frac{55}{221}$.
• B. 0
• C. Cant be determined
• D. None of the above

Answer 1

Answer: (A)

$\frac{55}{221}$.

Out of 52 cards, two cards can be drawn in ${}^{52}{C}_2$ ways.
So, total number of elementary events ${=}^{52}{C}_2$
Consider the following events:
$A=$ Two cards drawn are red cards;
$B=$ Two cards drawn are kings.
Required probability
$=P(A\cup B)$
$=P\left(A\right)+P\left(B\right)-P(A\cap B)....\left(i\right)$
Now we shall find P(A), P(B) and $P(A\cap B)$
There are 26 red cards, out of which 2 red cards can be drawn in ${}^{26}{C}_2$ ways.
$\therefore P\left(A\right)=\frac{{}^{26}{C}_2}{{}^{52}{C}_2}$
Since there are 4 kings, out of which 2 kings can be drawn in ${}^4{C}_2$ ways.
$\therefore P\left(B\right)=\frac{{}^4{C}_2}{{}^{52}{C}_2}$
There are 2 cards which are both red and kings. Therefore, $P(A\cap B)=$ Probability of getting 2 cards which are both red and kings. $=$ Probability of getting 2 red kings.
$=\frac{{}^2{C}_2}{{}^{52}{C}_2}$
Now,
Required probability
$=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$=\frac{{}^{26}{C}_2}{{}^{52}{C}_2}+\frac{{}^4{C}_2}{{}^{52}{C}_2}-\frac{{}^2{C}_2}{{}^{52}{C}_2}$
$=\frac{325}{1326}+\frac{1}{221}-\frac{1}{1326}=\frac{55}{221}$