Match List - I with List - II. Choose the correct answer from the options given below :List - I (Queries) List - II (Probability) (A) A bag contains 6 white and 4 red balls. Two balls are drawn at random. What is the chance, they will be the same colour ? (I) 3/68 (B) In a pack of 52 cards, one card is drawn at random, what is the probability that it is either a king or a queen ? (II) 14/68 (C) A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability of 1 red and 2 white balls ? (III) 2/13 (D) A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random. Find the probability of 2 blue and 1 red balls ? (IV) 7/15
(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Each probability is (favourable selections) ÷ (total selections), counting unordered selections with combinations \(\binom{n}{r}\).
Step 1 — (A): 10 balls, choose 2: \(\binom{10}{2} = 45\). Same colour = both white or both red: \(\binom{6}{2} + \binom{4}{2} = 15 + 6 = 21\). Probability \(\dfrac{21}{45} = \dfrac{7}{15}\), entry (IV).
Step 2 — (C) and (D): 18 balls, choose 3: \(\binom{18}{3} = 816\). One red and two white: \(6 \times \binom{4}{2} = 36\), so \(\dfrac{36}{816} = \dfrac{3}{68}\), entry (I). Two blue and one red: \(\binom{8}{2} \times 6 = 168\), so \(\dfrac{168}{816} = \dfrac{14}{68}\), entry (II).
Step 3 — (B): there are 4 kings and 4 queens, so the probability is \(\dfrac{8}{52} = \dfrac{2}{13}\). This is exactly entry (III).
The matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II). Codes that send (A) to (III) fail because 7/15 belongs to the same-colour query, and codes that swap (C) and (D) give 14/68 to the white-ball case, which is only 3/68.
(A)-(IV), (B)-(III), (C)-(I), (D)-(II) option is correct.
Similar Questions
Consider an experiment of tossing two fair dice, one black and one red. What is the probability that the number on the black die divides the number on red die ?
A coin is tossed successively three times. Find the Probability (P), Event (E), Sample space (S) of getting exactly one head or two heads, where n is number of occurrence.
(A) n(S) = 8 and n(E) = 4
(B) n(E) = 6 and n(S) = 8
(C) P(E) = \(\dfrac{3}{4}\)
(D) P(E) = \(\dfrac{1}{2}\)
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