Let ‘S’ be the sample space. Then, n(S) = 52.
Let A, B and C be the events of getting a king, a heart and a red card respectively.
Then, n(A) = 4, n(B) = 13, n(C) = 26
∴ P(A) = \(\frac{n(A)}{n(S)}= \frac{4}{52}\)
P(B) = \(\frac{n(B)}{n(S)}= \frac{13}{52}\)
P(C) = \(\frac{n(C)}{n(S)}=\frac{26}{52}\)
n(A ∩ B) = 1, n(B ∩ C) = 13, n(C ∩ A) = 2, n(A ∩ B ∩ C) = 1
⇒ P(A ∩ b) = \(\frac{1}{52}\) ⇒ P(B ∩ C) = \(\frac{13}{52}\)
⇒ P(A ∩ B) = \(\frac{1}{52}\) ⇒ n(A ∩ B ∩ C) = \(\frac{1}{52}\)
We know–
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)
= \(\frac{4}{52}+\frac{13}{52}+\frac{26}{52}-\frac{1}{52}-\frac{13}{52}-\frac{2}{52}+\frac{1}{52}\)
= \(\frac{28}{52}\)
= \(\frac{7}{13}\)
Thus, the probability of getting a king, a heart of a red card is 7/13.
