Question

12. In testing a certain kind of truck tire over rugged terrain, it is found that 25% of the trucks fail to complete the test run without a blowout. Of the next 15 trucks tested, find the probability that

(a) from 3 to 6 have blowouts;

(b) fewer than 4 have blowouts;

(c) more than 5 have blowouts

Answer 1

It is given information that in testing a certain kind of truck tire over a rugged terrain, it is found that 25% of the trucks fail to complete the test run without a blowout.

Let X denote the random variable representing the number of trucks having blowouts during the test run out of next 15 trucks tested.

Let us consider a truck having a blowout during the test run as a success.

Therefore, probability of a success in each trial is p = 0.25

n = 15

Since the trials are independent so, X ~ Bin(n,p)

The probability mass function (p.m.f) is

P (X=x) = b(x;15,0.25), where x = 0, 1, 2,…,15

$= \binom{15}{x}(0.25)^x(1 -0.25)^{15-x}$

$= \binom{15}{x}(0.25)^x(0.75)^{15-x}$ , where x = 0, 1, 2,…,15

(a) Compute the probability that the next 15 trucks tested from 3 to 6 have blowouts.

$P(3≤X≤6) = P(X=3) + P(X=4) + P(X=5) + P(X=6) \\ = \binom{15}{3}(0.25)^3(0.75)^{15-3} + \binom{15}{4}(0.25)^4(0.75)^{15-4} + \binom{15}{5}(0.25)^5(0.75)^{15-5} + \binom{15}{6}(0.25)^6(0.75)^{15-6} \\ = 0.2252 + 0.2252 + 0.1651 + 0.0917 \\ = 0.707$

Therefore, the probability that the next 15 trucks tested from 3 to 6 have blowouts is 0.707.

(b) Compute the probability that the next 15 trucks tested fewer than 4 have blowouts.

$P(X<4) = P(X≤3) \\ = P(X=0) + P(X=1) + P(X=2) + P(X=3) \\ = \binom{15}{0}(0.25)^0(0.75)^{15-0} + \binom{15}{1}(0.25)^1(0.75)^{15-1} \binom{15}{2}(0.25)^2(0.75)^{15-2} + \binom{15}{3}(0.25)^3(0.75)^{15-3} \\ = 0.0134 + 0.0668 + 0.1559 + 0.2252 \\ = 0.4613$

Therefore, the probability that the next 15 trucks tested fewer than 4 have blowouts is 0.4613.

(c) Compute the probability that the next 15 trucks tested more than 5 have blowouts.

$P(X>5) = 1 -P(X≤5) \\ = 1 -(P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)) \\ = 1 -( \binom{15}{0}(0.25)^0(0.75)^{15-0} + \binom{15}{1}(0.25)^1(0.75)^{15-1} \binom{15}{2}(0.25)^2(0.75)^{15-2} + \binom{15}{3}(0.25)^3(0.75)^{15-3} + \binom{15}{4}(0.25)^4(0.75)^{15-4} + \binom{15}{5}(0.25)^5(0.75)^{15-5} ) \\ = 1 -( 0.0134 + 0.0668 + 0.1559 + 0.2252 + 0.2252 + 0.1651 ) \\ = 0.1484$

Therefore, the probability that the next 15 trucks tested more than 5 have blowouts is 0.1484.