Maharashtra State BoardHSC Arts 12th Board Exam
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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 - Mathematics and Statistics

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Sum

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

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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` 

Concept: Binomial Distribution
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 8 Binomial Distribution
Miscellaneous exercise 8 | Q 5.2 | Page 254
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