(d) \(\frac{7}{13}\)
Here n(S) = 52
Let A, B, C be the events of getting a red card, a diamond and a jack respectively.
∵ There are 26 red cards, 13 diamonds and 4 jacks, n(A) = 26, n(B) = 13, n(C) = 4
⇒ n(A ∩ B) = n(red and diamond) = 13,
n(B ∩ C) = n(diamond and jack) = 1,
(∵ There is only jack of diamonds)
n(A ∩ C) = n(red and jack) = 2
(∵ There are two red jacks)
n(A ∩ B ∩ C) = n(red, diamond, jack) = 1
∴ P(A) = \(\frac{26}{52}\), P(B) = \(\frac{13}{52}\), P(C) = \(\frac{4}{52}\), P(A\(\cap\)B) = \(\frac{13}{52}\),
P(B \(\cap\)C) = \(\frac{1}{52}\), P(A\(\cap\)C) = \(\frac{2}{52}\), P(A\(\cap\)B\(\cap\)C) = \(\frac{1}{52}\)
∴ P(a red card or a diamond or a jack) = P(A ∪ B ∪ C)
= P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)
= \(\frac{26}{52}\) + \(\frac{13}{52}\) + \(\frac{4}{52}\) - \(\bigg(\)\(\frac{13}{52}\) + \(\frac{1}{52}\) + \(\frac{2}{52}\)\(\bigg)\) + \(\frac{1}{52}\)
= \(\frac{44}{52}\) + \(\frac{16}{52}\) = \(\frac{28}{52}\) = \(\frac{7}{13}\) .