(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)