Chapter 13.2, Problem 54AYU

Selecting Objects An urn contains 15 red balls and 10 white balls. Five balls are selected. In how many ways can the 5 balls be drawn from the total of 25 balls:

(a) If all 5 balls are red?

(b) If 3 balls are red and 2 are white?

(c) If at least 4 are red balls?

To determine

To find: In how many ways can the 5 balls be drawn from the total of 25 balls:

a. If all 5 balls are red?

Answer to Problem 54AYU

a. 3003

Explanation of Solution

Given:

An urn contains 15 red balls and 10 white balls. Five balls are selected. 5 balls can be drawn from the total of 25 balls.

Formula used:

$C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

Number of red balls-15.

Number of white balls-10.

Total number of balls -25.

a. Number of ways 5 balls is drawn such that all of them are red.

$=C\left(15,5\right)*C\left(10,0\right)$

$\frac{15!}{10!5!}*\frac{10!}{10!0!}$

$3003*1$

$3003$

To determine

To find: In how many ways can the 5 balls be drawn from the total of 25 balls:

b. If 3 balls are red and 2 are white?

Answer to Problem 54AYU

b. 20475

Explanation of Solution

Given:

An urn contains 15 red balls and 10 white balls. Five balls are selected. 5 balls can be drawn from the total of 25 balls.

Formula used:

$C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

Number of red balls-15.

Number of white balls-10.

Total number of balls -25.

b. Number of ways 5 balls is drawn such that 3 balls are red and 2 balls are white.

$=C\left(15,3\right)*C\left(10,2\right)$

$\frac{15!}{3!12!}*\frac{10!}{2!8!}$

$455*45$

$20475$

To determine

To find: In how many ways can the 5 balls be drawn from the total of 25 balls:

c. If at least 4 are red balls?

Answer to Problem 54AYU

c. 16653

Explanation of Solution

Given:

An urn contains 15 red balls and 10 white balls. Five balls are selected. 5 balls can be drawn from the total of 25 balls.

Formula used:

$C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

Number of red balls-15.

Number of white balls-10.

Total number of balls -25.

c. Number of ways 5 balls is drawn such that at least 4 balls are red.

$4\text{ balls are red and }1\text{ white ball }+ 5\text{ balls are red}$

$=C\left(15,4\right)*C\left(10,1\right)+C\left(15,5\right)*C\left(10,10\right)$

$\frac{15!}{4!11!}*\frac{10!}{1!10!}+\frac{15!}{5!10!}*1$

$1365*10+3003*1$

$13650+3003$

$16653$