Question

this quiz will test your knowledge on permutations and combinations. value: 1 a shipment of thirteen smartphones contains eight with cracked screens. if sold in a random order, what is the probability that exactly three of the first five sold have cracked screens? a. 44% b. 35% c. 47% d. 38%

Explanation:

Step1: Calculate number of ways to choose 3 cracked - screen phones out of 8

Use combination formula $C(n,k)=\frac{n!}{k!(n - k)!}$, where $n = 8$ and $k=3$.
$C(8,3)=\frac{8!}{3!(8 - 3)!}=\frac{8!}{3!5!}=\frac{8\times7\times6}{3\times2\times1}=56$

Step2: Calculate number of ways to choose 2 non - cracked - screen phones out of 5

There are $13 - 8=5$ non - cracked - screen phones. Using combination formula with $n = 5$ and $k = 2$.
$C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5!}{2!3!}=\frac{5\times4}{2\times1}=10$

Step3: Calculate number of ways to choose 5 phones out of 13

Using combination formula with $n = 13$ and $k = 5$.
$C(13,5)=\frac{13!}{5!(13 - 5)!}=\frac{13\times12\times11\times10\times9}{5\times4\times3\times2\times1}=1287$

Step4: Calculate the probability

The probability $P$ that exactly 3 of the first 5 sold have cracked screens is the product of the number of ways to choose 3 cracked - screen and 2 non - cracked - screen phones divided by the number of ways to choose 5 phones out of 13.
$P=\frac{C(8,3)\times C(5,2)}{C(13,5)}=\frac{56\times10}{1287}\approx0.435$ or $44\%$

Answer:

A. 44%