From a well shuffled pack of $52$ playing cards two cards drawn at random. The probability that either both are red or both are kings is:

\frac{(^{26} C_{2} +^{4} C_{2})}{^{52} C_{2}}

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\frac{(^{26} C_{2} +^{4} C_{2} -^{2} C_{2})}{^{52} C_{2}}

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\frac{^{30} C_{2}}{^{52} C_{2}}

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\frac{^{39} C_{2}}{^{52} C_{2}}

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Solution

The correct option is B $\frac{(^{26} C_{2} +^{4} C_{2} -^{2} C_{2})}{^{52} C_{2}}$
Assuming, cards are drawn without replacement:

Total possible events $=^{52} C_{2}$

P(both red) $= \frac{^{26} C_{2}}{^{52} C_{2}}$

P(both king) $= \frac{^{4} C_{2}}{^{52} C_{2}}$

P(both red & king) $= \frac{^{2} C_{2}}{^{52} C_{2}}$

We use the basic addition rule,

P(both black or both queens) = P(both red) + P(both queens) - P(both black as well as queens)

Required probability $= \frac{^{26} C_{2} +^{4} C_{2} -^{2} C_{2}}{^{52} C_{2}}$

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