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in Probability and Probability Distribution by (46.2k points)

An urn contains 5 white, 7 red and 8 black balls. If 4 balls are drawn one by one with replacement, what is the probability that: 

(i) all are white 

(ii) only 3 are white 

(iii) none is white 

(iv) at least 3 are white

1 Answer

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Best answer

(i) Total number of balls 

= 5 + 7 + 8 = 20 

Number of white balls = 5 

Probability of getting white ball is one chance = 5/20 = 1/4

∵ All events are independent 

∴ Required probability = 1/4 × 1/4 × 1/4 × 1/4 = (1/4)

(ii) Probability of drawing white ball first time 

= 3C1 ×1/4 × 1/4 × 1/4 

= 3 ×(1/4)3 

(iii) P(no ball is white) 

∴ Number of other balls = 7 + 8 = 15 

∴ Probability of drawing one other colour ball = 15/20 = 3/4

∴ Probability of other colour balls drawn successively (none is white) 

= 3/4 × 3/4 × 3/4 × 3/4 

= (3/4)4 

(iv) P(at least 3 white) = P (four white) + P(three white)

= (1/4)4 + 3/43 = 1/44 = 3/43

= 13/44

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