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

The probability questions involve calculating the chances of drawing two or more red balls, exactly two white balls and one red ball, and the latter given a white ball is drawn first from a box with five red and three white balls without replacement.

Step-by-step explanation:

To solve these probability problems, we need to calculate the odds of different combinations when drawing balls from a box without replacement. Given there are five red balls and three white balls, let's answer the questions:

• Probability of two or more red balls: We need to find the probability of drawing 3 red balls, or 2 red balls and 1 white ball. We calculate these probabilities separately and then sum them up.
• Probability of exactly two white balls and one red ball: This scenario requires us to find the probability of drawing these specific balls in any order.
• Probability of drawing two white balls and one red ball, given a white is drawn first: We adjust our sample space since we know the first ball is white, and then calculate the probability.

These problems are similar to other probability questions involving drawing cards from a deck or marbles from a bag, but with the unique condition that the balls are not replaced, affecting the probabilities for subsequent draws.

Answer 2

a.
`(63)/(128)`

b.
`(135)/(512)`

c.
`(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 ( (5)/(8) \right )^2\left ( (3)/(8) \right )+\left ( (3)/(8) \right )^3=(225)/(512)+(27)/(512)=(252)/(512)=(126)/(256)=(63)/(128)`

b. Probability that exactly two white balls and one red ball will be obtained in the three draws =
`3\left ( (5)/(8) \right )\left ( (3)/(8) \right )^2=(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 ( (2)/(7) \right )^2\left ( (5)/(7) \right )=(60)/(343)`