Question

An urn contains 10 red and 15 blue balls.

(a) (5 points) You select 8 balls

at random without replacement. What is the probability that

exactly 5 are blue?

(b) (5 points) You select 3 balls

at random without replacement. What is the probability that

exactly two are the same color?

(c) You select 3 balls

at random without replacement. You find that exactly two are the same color.

What is the conditional probability that the two that are the same color are both blue?

Answer 1

(a) An urn contains $10$ red and $15$ blue balls. Out of which $8$ balls selected at random.

From the total of $25$ balls , $8$ balls can be select in $^{25}C_8$ ways.

Therefore sample space $S$ contains $^{25}C_8$ number of elements.

Let $A$ be an event such that, $A=$ Exactly $5$ balls are blue.

Then exactly $5$ blue balls can be select in $^{15}C_5×^{10}C_3$ ways.

Number of elements in favour of $A$ is $^{15}C_5×^{10}C_3$ .

$\therefore P(A)= \frac{n(A)}{n(S)}=\frac{^{15}C_5×^{10}C_3}{^{25}C_8}$ $=\frac{3003×120}{1081575}=\frac{8008}{24205}$

(b) From the total of $25$ balls $3$ balls can be select in $^{25}C_3$ ways.

So sample space $S$ contains $^{25}C_3$ number of elements.

Let $B$ be an event such that, $B=$ Exactly $2$ balls are same color

Now out of $3$ balls $2$ same color balls can be select in two ways. Either it will be $2$ red balls and $1$ blue ball or it will be $2$

blue balls and $1$ blue balls.

So for the first case $3$ balls can be select in $^{10}C_2×^{15}C_1$ ways . And for the second case $3$ balls can be select in $^{10}C_1×^{15}C_2$ ways.

Therefore number of elements in favour of the event $A$ is $[^{10}C_2×^{15}C_1+^{10}C_1×^{15}C_2]$

$\therefore$ $P(B)=\frac {[^{10}C_2×^{15}C_1+^{10}C_1×^{15}C_2]}{^{25}C_3}$ = $\frac{1725}{1081575}=\frac{23}{24035}$

(c) Let $B$ and $C$ are two events such that,

$B=$ Exactly $2$ balls are same color

$C=$ Two same color balls are blue

From the total of $25$ balls $3$ balls can be select in $^{25}C_3$ ways.

So sample space $S$ contains $^{25}C_3$ number of elements.

$\therefore P(B)=\frac {[^{10}C_2×^{15}C_1+^{10}C_1×^{15}C_2]}{^{25}C_3}$ $=\frac{1725}{^{25}C_3}$

And $P(B\cap C)=\frac{^{10}C_1×^{15}C_2}{^{25}C_3}=\frac{1050}{^{25}C_3}$

So the required probability is $P(C/B)=\frac{P(B\cap C)}{P(B)}=\frac{\frac{1050}{^{25}C_3}}{\frac{1725}{^{25}C_3}}=\frac{1050}{1725}=\frac{14}{23}$