Practice | Hashtable

Home > Practice > Arithmetic Aptitude > Probability > Miscellaneous

1. Two cards are drawn from a pack of 52 cards. The probability that either both are red or both are king, is-

A. $\frac{7}{13}$

B. $\frac{3}{26}$

C. $\frac{63}{221}$

D. $\frac{55}{221}$

Solution:

Clearly,
n (S) =

n (S) =

^{52} C_{2} =

\frac{(52 \times 51)}{2}

= 1326
Let

E_{1}

= event of getting both red cards

E_{2}

= event of getting both kings
Then,

E_{1}

\cap

E_{2}

= event of getting 2 kings of red cards.
∴

n (E_{1}) =^{26} C_{2} = \frac{(26 \times 25)}{(2 \times 1)}

= 325 and

n (E_{2}) =^{4} C_{2} = \frac{(4 \times 3)}{(2 \times 1)}

= 6

n (E_{1} \cap E_{2}) =^{2} C_{2} = 1

∴ P (E_{1}) = \frac{n (E_{1})}{n (S)} = \frac{325}{1326}

and

P (E_{2}) = \frac{n (E_{2})}{n (S)} = \frac{6}{1326}

P (E_{1} \cap E_{2}) = \frac{1}{1326}

∴ P (both red or both kings)

\begin{aligned} = P (E_{1} \cup E_{2}) \\ = P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2}) \\ = (\frac{325}{1326} + \frac{6}{1326} - \frac{1}{1326}) \\ = \frac{330}{1326} \\ = \frac{55}{221} \end{aligned}

You must login to add comments. Login now.