Question

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

Number of sample points ⁵²C₂​

Let 'A' be the event two cards drawn are red 

∴ n(A) = ²⁶C_2​

'B' be the event two cards drawn are kings

∴ n(B) = ⁴C_2​

∴ P(A or B) = P(A∪B) = P(A) + P(B) − P(A∩B)  .....(1)

where P(A) = \(\frac{^{26}C_2}{^{52}C_2} = \frac{325}{1326}\)

P(B) = \(\frac{^4C_2}{^{52}C_2} = \frac{1}{221}\) and

P(A∩B) = \(\frac{^2C_2}{^{52}C_2} = \frac{1}{1326}\)

Using (1) we have,

P(A∪B) = \( \frac{325}{1326} + \frac 1{221} - \frac 1{1326}= \frac{55}{221}\)

Answer 2

Let S : Drawing 2 cards out of 52 card 

A : Drawing 2 red cards 

B : Drawing 2 kings 

A ∪ B : Drawing 2 red cards or 2 kings 

∴ n(S) = ⁵²C₂ 

n(A) = ²⁶C₂          (∵ There are 26 red cards) 

n(B) = ⁴C₂          (∵ There are 4 kings) 

But there are 2 red kings, so 

A ∩ B : Drawing 2 red kings 

⇒ n(A ∩ B) = ²C₂. 

∴ Required probability = P(A∪B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{n(A)}{n(S)}\) + \(\frac{n(B)}{n(S)}\) - \(\frac{n(A\cap{B})}{n(S)}\) = \(\frac{^{26}C_2}{^{52}C_2}\) + \(\frac{^{4}C_2}{^{52}C_2}\) - \(\frac{^{2}C_2}{^{52}C_2}\)

= \(\frac{26\times25}{52\times51}\) + \(\frac{4\times3}{52\times51}\) - \(\frac{2}{52\times51}\) = \(\frac{660}{2652}\) = \(\frac{55}{221}.\)