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The probability that a bulb produced by a factory will fuse after 6 months of use is 0.05. find the probability that out of 5 such bulbs

(i) none will fuse after 6 months of use

(ii) at least one will fuse after 6 months of use

(iii) not more than one will fuse after 6 months of use

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(i) The probability that the bulb will fuse = 0.05 = p

The probability that the bulb will not fuse = 1-0.05 = 0.95 = q

Using Bernoulli’s we have,

P(Success=x) \(=n_{C_x}.p^x.q^{(n-x)}\)

x = 0, 1, 2, ………n and q = (1 - p), n = 5

Probability that none will fuse

\(=5_{C_0}.(0.05)^0(0.95)^5\)

\(\Rightarrow (0.95)^5\)

(ii) The probability that the bulb will fuse = 0.05 = p

The probability that the bulb will not fuse = 1-0.05 = 0.95 = q

Using Bernoulli’s we have,

P(Success=x) \(=n_{C_x}.p^x.q^{(n-x)}\)

x = 0, 1, 2, ………n and q = (1 - p), n = 5

Probability that at least one will fuse = P(1) + P(2) + P(3) + P(4) + P(5)

  

(iii) The probability that the bulb will fuse = 0.05 = p

The probability that the bulb will not fuse = 1-0.05 = 0.95 = q

Using Bernoulli’s we have,

P(Success=x) \(=n_{C_x}.p^x.q^{(n-x)}\)

x = 0, 1, 2, ………n and q = (1 - p), n = 5

Probability that not more than one will fuse = P(0) + P(1)

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