Question

One card is drawn from each of two ordinary sets of 52 cards.Find the probability that at least one of them will be Ace of hearts.

• A. $\frac{1}{52}$
• B. $\frac{1}{26}$
• C. $\frac{103}{2704}$
• D. $\frac{824}{2704}$

Answer 1

Answer: (C)

$\frac{103}{2704}$

Let A be the event that the first card is the ace of hearts and B the corresponding event in respect of the second card.

We have to find $P(A\cup B).$

But by addition theorem, $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)...\left(1\right)$

Now $P\left(A\right)=P\left(B\right)=\frac{1}{52},$

Since there is only ace of hearts in each suit of 52 cards.

To find $P(A\cap B),$ we see that there are in all ${52}^2$ cases which are represented by all possible pairs consisting of one card from first and one from the second. Only one of these cases is favourable to the event $A\cap B$

So $P(A\cap B)=\frac{1}{{52}^2}.$

Hence from (1); we obtain $P(A\cup B)=\frac{1}{52}+\frac{1}{52}-\frac{1}{{52}^2}=\frac{103}{2704}.$