Question

ten cars were randomly assembled at a factory. of these cars, 3 will be chosen for inspection. of the 10 cars, 4 are blue and 6 are grey. what is the probability that all 3 cars picked for inspection will be grey?

Explanation:

Step1: Calculate total number of ways to pick 3 cars

The number of ways to choose 3 cars out of 10 is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 10$ and $r=3$. So $C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1}=120$.

Step2: Calculate number of ways to pick 3 grey cars

The number of ways to choose 3 grey cars out of 6 is given by the combination formula with $n = 6$ and $r = 3$. So $C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6\times5\times4}{3\times2\times1}=20$.

Step3: Calculate the probability

The probability $P$ that all 3 cars picked are grey is the number of favorable outcomes (picking 3 grey cars) divided by the number of total outcomes (picking any 3 cars). So $P=\frac{C(6,3)}{C(10,3)}=\frac{20}{120}=\frac{1}{6}$.

Answer:

$\frac{1}{6}$