Question

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. 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

Let n = 15, p = 25% = 0.25, q = 1 - p = 0.75

According to the binomial distribution:

$P(x=k)=C_n^kp^kq^{n-k}$

$a)\space P(3\leq x\leq6)=\sum_{k=3}^{6}P(x=k)=\\ =\sum_{k=3}^{6}C_n^kp^kq^{n-k}=\\ \frac{15\cdot14\cdot13}{1\cdot2\cdot3}0.25^30.75^{12}+\frac{15\cdot14\cdot13\cdot12}{1\cdot2\cdot3\cdot4}0.25^40.75^{11}+\\ +\frac{15\cdot14\cdot13\cdot12\cdot11}{1\cdot2\cdot3\cdot4\cdot5}0.25^50.75^{10}+\\ +\frac{15\cdot14\cdot13\cdot12\cdot11\cdot10}{1\cdot2\cdot3\cdot4\cdot5\cdot6}0.25^60.75^9=\\ =0.225199065+0.225199065+\\ +0.165145981+0.091747767=\\ =0.707291878\approx70.73\%$

$b)\space P(x\lt4)=\sum_{k=0}^{3}P(x=k)=\\ =\sum_{k=0}^{3}C_n^kp^kq^{n-k}=\\ 0.25^00.75^{15}+\frac{15}{1}0.25^10.75^{14}+\frac{15\cdot14}{1\cdot2}0.25^20.75^{13}+\\ +\frac{15\cdot14\cdot13}{1\cdot2\cdot3}0.25^30.75^{12}=\\ =0.013363461+0.066817305+\\ +0.155907045+0.225199065=\\ =0.461286876\approx46.13\%$

$c)\space P(x\gt5)=\sum_{k=6}^{15}C_n^kp^kq^{n-k}=\\ (p+q)^n-\sum_{k=0}^{5}C_n^kp^kq^{n-k}=\\ 1-0.461286876-\frac{15\cdot14\cdot13\cdot12}{1\cdot2\cdot3\cdot4}0.25^40.75^{11}+\\ +\frac{15\cdot14\cdot13\cdot12\cdot11}{1\cdot2\cdot3\cdot4\cdot5}0.25^50.75^{10}=\\ =0.538713124-0.225199065-0.165145981=\\ =0.148368078\approx 14.84\%$