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InterviewSolution
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A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is (i) a black king (ii) either a black card or a king (iii) black and a king (iv) a jack, queen or a king (v) neither a heart nor a king (vi) spade or an ace (vii) neither an ace nor a king (viii) neither a red card nor a queen. (ix) other than an ace (x) a ten (xi) a spade (xii) a black card (xiii) the seven of clubs (xiv) jack (xv) the ace of spades (xvi) a queen (xvii) a heart (xviii) a red card (xix) neither a king nor a queen |
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Answer» Total number of outcomes, n(S) = 52 (i) n(E) = 2 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\) (ii) n(E) = 26 + 2 = 28 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{28}{52}\) = \(\frac{7}{13}\) (iii) n(E) = 2 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{2}{52}\) = \(\frac{1}{26}\) (iv) n(E) = 12 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{12}{52}\) = \(\frac{3}{13}\) (v) n(E) = 36 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{36}{52}\) = \(\frac{9}{13}\) (vi) n(E) = 16 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\) (vii) n(E) = 44 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{44}{52}\) (viii) n(E) = 24 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{24}{52}\) = \(\frac{6}{13}\) (ix) n(E) = 48 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{48}{52}\) = \(\frac{12}{13}\) (x) n(E) = 4 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{52}\) = \(\frac{1}{13}\) (xi) n(E) = 13 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{13}{52}\) = \(\frac{1}{4}\) (xii) n(E) = 26 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{26}{52}\) = \(\frac{1}{2}\) (xiii) n(E) = 1 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{1}{52}\) (xiv) n(E) = 4 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{52}\) = \(\frac{1}{13}\) (xv) n(E) = 1 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{1}{52}\) (xvi) n(E) = 4 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{52}\) = \(\frac{1}{13}\) (xvii) n(E) = 13 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{13}{52}\) = \(\frac{1}{4}\) (xviii) n(E) = 26 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{26}{52}\) = \(\frac{1}{2}\) (xix) n(E) = 44 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{44}{52}\) = \(\frac{11}{13}\) |
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