7) (6 points) a bag contains 15 balls 8 of which are red and the rest are green. 4 balls are drawn from the…

Explanation:

Step1: Calculate total number of ways to draw 4 balls

The total number of ways to draw 4 balls out of 15 is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 15$ and $r=4$. So, $C(15,4)=\frac{15!}{4!(15 - 4)!}=\frac{15\times14\times13\times12}{4\times3\times2\times1}=1365$.

Step2a: Calculate number of ways to draw 4 red balls

There are 8 red balls. The number of ways to draw 4 red balls out of 8 is $C(8,4)=\frac{8!}{4!(8 - 4)!}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70$. The probability that all 4 drawn balls are red is $\frac{C(8,4)}{C(15,4)}=\frac{70}{1365}=\frac{2}{39}$.

Step2b: Calculate number of ways to draw 2 red and 2 green balls

There are 8 red balls and $15 - 8=7$ green balls. The number of ways to draw 2 red balls out of 8 is $C(8,2)=\frac{8!}{2!(8 - 2)!}=\frac{8\times7}{2\times1}=28$. The number of ways to draw 2 green balls out of 7 is $C(7,2)=\frac{7!}{2!(7 - 2)!}=\frac{7\times6}{2\times1}=21$. The number of ways to draw 2 red and 2 green balls is $C(8,2)\times C(7,2)=28\times21 = 588$. The probability that 2 are red and 2 are green is $\frac{C(8,2)\times C(7,2)}{C(15,4)}=\frac{588}{1365}=\frac{84}{195}=\frac{28}{65}$.

Step2c: Calculate number of ways to draw 4 green balls

The number of ways to draw 4 green balls out of 7 is $C(7,4)=\frac{7!}{4!(7 - 4)!}=\frac{7\times6\times5}{3\times2\times1}=35$. The probability that all 4 drawn balls are green is $\frac{C(7,4)}{C(15,4)}=\frac{35}{1365}=\frac{1}{39}$. The probability that all drawn balls are of the same color is the sum of the probability of all - red and all - green cases, which is $\frac{2}{39}+\frac{1}{39}=\frac{3}{39}=\frac{1}{13}$.

Answer:

a) $\frac{2}{39}$ b) $\frac{28}{65}$ c) $\frac{1}{13}$