The probability that a mountain-bike travelling along a certain track will have a tyre burst is 0.05. Find the probability that among 17 riders: at most three have a burst tyre
Solution
Let X = number of burst tyre.
p = probability that a mountain-bike travelling along a certain track will have a tyre burst
∴ p = 0.05
∴ q = 1 - p = 1 - 0.05 = 0.95
Given: n = 17
∴ X ~ B(17, 0.05)
The p.m.f. of X is given by
P(X = x) = `"^nC_x p^x q^(n - x)`
i.e. p(x) = `"^17C_x (0.05)^x (0.95)^(17 - x)`, x = 0, 1, 2,...,17
P (at most three have a burst tyre) = P(X ≤ 3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= p(0) + p(1) + p(2) + p(3)
`= ""^17C_0 (0.05)^0(0.95)^(17-0) + ""^17C_1 (0.05)^1 (0.17)^(17 - 1) + ""^17C_2 (0.05)^2 (0.17)^(17 - 2) + "^17C_3 (0.05)^3 (0.17)^(17 - 3)`
`= 1(1)(0.95)^17 + 17(0.05)(0.95)^16 + (17 xx 16)/(2 xx 1) xx (0.05)^2 (0.95)^15 + (17 xx 16 xx 15)/(3 xx 2 xx 1) xx (0.05)^3 xx (0.95)^14`
`= (0.95)^17 + 17(0.05) xx (0.95)^16 + 17(8) xx (0.05)^2 xx (0.95)^15 + 17(8)(5) xx (0.05)^3 xx (0.95)^14`
`= (0.95)^14 [(0.95)^3 + (17)(0.05)(0.95)^2 + 17(8) xx (0.05)^2 xx (0.95)^1 + 17(8)(5)(0.05)^3]`
`= (0.95)^14 [0.8574 + 0.7671 + 0.323 + 0.085]`
`= (2.0325)(0.95)^14`
Hence, the probability that at most three riders have burst tyre `= (2.0325)(0.95)^14`
