Question

There are five red balls and three white balls in a box. An experiment consist of drawing three balls in succession. Once a ball is drawn, it is not replaced.

a. What is the probability that two or more red balls will be obtained in the three draws?
b. What is the probability that exactly two white balls and one red ball will be obtained in the three draws?
c. Given a white ball is selected on the first draw, what is the probability of selecting two white balls and one red ball after the three draws?

Answer 1

Answer:

a. $\frac{63}{128}$

b. $\frac{135}{512}$

c. $\frac{60}{343}$

Step-by-step explanation:

Probability refers to chances of occurrence of some event.

Let R denotes the event: ball drawn is red

Let W denotes the event: ball drawn is white

a. Probability that two or more red balls will be obtained in the three draws = probability that two red balls will be obtained in the three draws + probability that three red balls will be obtained in the three draws

= $3\left ( \frac{5}{8} \right )^2\left ( \frac{3}{8} \right )+\left ( \frac{3}{8} \right )^3=\frac{225}{512}+\frac{27}{512}=\frac{252}{512}=\frac{126}{256}=\frac{63}{128}$

b. Probability that exactly two white balls and one red ball will be obtained in the three draws = $3\left ( \frac{5}{8} \right )\left ( \frac{3}{8} \right )^2=\frac{135}{512}$

c. As ball selected on the first draw is white, number of white balls left = 2

Number of red balls = 5

Probability of selecting two white balls and one red ball after the three draws = $3\left ( \frac{2}{7} \right )^2\left ( \frac{5}{7} \right )=\frac{60}{343}$