Answer to Question #198625 in Statistics and Probability for Tashmi

Question

9. A large company that must hire a new president prepares a final list of five candidates, all

of whom are qualified. Two of these candidates are members of a minority group. To

avoid bias in the selection of the candidate, the company decides to select the president

by lottery. What is the probability that one of the minority candidates is hired?

10. On February 1, 2003, the space shuttle Columbia exploded. This was the second disaster

in 113 space missions for NASA. On the basis of this information, what is the probability

that a future mission is successfully completed?

11. What is the probability that a card chosen at random from a standard deck of cars will be

either a king or a heart?

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

9.

$p=\frac{2}{5}\cdot\frac{3}{4}=\frac{3}{10}$

10.

$p=\frac{113-2}{113}=\frac{111}{113}$

11.

$p=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}$

12.

Binomial probability:

$P(x=k)=C^k_np^k(1-p)^{n-k}$

$p=0.25,n=15$

a)

$P(3\le x\le6)=P(x=3)+P(x=4)+P(x=5)+P(x=6)$

$P(x=3)=C^3_{15}\cdot0.25^3\cdot0.75^{12}=455\cdot0.25^3\cdot0.75^{12}=0.2252$

$P(x=4)=C^4_{15}\cdot0.25^4\cdot0.75^{11}=1365\cdot0.25^4\cdot0.75^{11}=0.2252$

$P(x=5)=C^5_{15}\cdot0.25^5\cdot0.75^{10}=3003\cdot0.25^5\cdot0.75^{10}=0.1651$

$P(x=6)=C^6_{15}\cdot0.25^6\cdot0.75^{9}=5005\cdot0.25^6\cdot0.75^{9}=0.0917$

$P(3\le x\le6)=0.2252+0.2252+0.1651+0.0917=0.7072$

b)

$P(x<4)=P(x=0)+P(x=1)+P(x=2)+P(x=3)$

$P(x=0)=0.75^{15}=0.0134$

$P(x=1)=15\cdot0.25\cdot0.75^{14}=0.0668$

$P(x=2)=C^2_{15}\cdot0.25^2\cdot0.75^{13}=105\cdot0.25^2\cdot0.75^{13}=0.1559$

$P(x<4)=0.0134+0.0668+0.1559+0.2252=0.4613$

c)

$P(x>5)=1-P(x\le5)=$

$=1-(P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5))$

$P(x>5)=1-(0.0134+0.0668+0.1559+0.2252+0.2252+0.1651)=0.1484$