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A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ?

(a) \(\frac{10}{13}\)

(b) \(\frac{15}{26}\)

(c) \(\frac{41}{52}\)

(d) \(\frac{7}{13}\)

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(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)

\(\frac{26}{52}\) + \(\frac{13}{52}\) + \(\frac{4}{52}\) - \(\bigg(\)\(\frac{13}{52}\) + \(\frac{1}{52}\) + \(\frac{2}{52}\)\(\bigg)\) + \(\frac{1}{52}\)

\(\frac{44}{52}\) + \(\frac{16}{52}\) = \(\frac{28}{52}\) = \(\frac{7}{13}\) .

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