The probability that a bulb produced by a factory will fuse after 150

1.	The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs. (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use
Answer» Probability that a bulb gets fuse after 150 days of its use = 0.05 Probability that the bulb will not fuse after 150 days of its use = 1 – 0.05 = 0.95 (i) Probability that no bulb will fuse after 150 days of its use = P(none) = (0.95)⁵ = 0.7738 = 0.77(approx) (ii) P(not more than one ) = O(0) + P(1) = (0.95)⁵ + ⁵C₁ x (0.95)⁴ x (0.05) = (0.95)⁵[0.95 + 5 x 0.05] = (0.95)⁴ x 1.2 (iii) P(more than one) = P(2) + P(3) + P(4) + P(5) = [P(0) + P(1) + P(2) + P(3) + P(4) + P(5)] - [P(0) + P(1)] = 1 - [P(0) + P(1)] = 1 - (0.95)⁴ x 1.2

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs. (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use

Answer»

Probability that a bulb gets fuse after 150 days of its use = 0.05 Probability that the bulb will not fuse after 150 days of its use = 1 – 0.05 = 0.95

(i) Probability that no bulb will fuse after 150

days of its use = P(none) = (0.95)⁵ = 0.7738

= 0.77(approx)

(ii) P(not more than one ) = O(0) + P(1)

= (0.95)⁵ + ⁵C₁ x (0.95)⁴ x (0.05)

= (0.95)⁵[0.95 + 5 x 0.05] = (0.95)⁴ x 1.2

(iii) P(more than one)

= P(2) + P(3) + P(4) + P(5)

= [P(0) + P(1) + P(2) + P(3) + P(4) + P(5)] - [P(0) + P(1)]

= 1 - [P(0) + P(1)] = 1 - (0.95)⁴ x 1.2

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