An urn contains 5 white, 7 red and 8 black balls. If 4 balls are drawn

1.	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
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)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

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

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)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

Discussion

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