Question

A random experiment consists of selecting two balls in succession from an urn containing two black balls and and one white ball. (a) Specify the sample space for this experiment. (b) Suppose that the experiment is modified so that the ball is immediately put back into the urn after the first selection. What is the sample space now? number of repetitions of the experiment in part a? In part b? outcome of the first draw in either of these experiments? (c) What is the relative frequency of the outcome (white, white) in a large (d) Does the outcome of the second draw fron the urn depend in any way on the outcome of the first draw in either of these experiments?

Answer 1

Answer:

(a) (B,B) (B,W) (W,B)

(b) (B,B) (B,W) (W,B) (WW)

(c) 1/9

(d) Yes it does in the experiment in part (a)

Explanation:

Black balls = 2

White Balls = 1

No. of balls selected = 2

(a) The balls are selected without replacement so, we can define the sample space as the kinds of possible outcomes:

1st ball is black and the 2nd ball is also black: (B, B)

1st ball is black and the 2nd ball is white: (B,W)

1st ball is white and the second ball is black: (W,B).

So, the sample space is: (B,B), (B,W), (W,B)

(b) Now, the balls are selected with replacement i.e. the first ball is taken out and then put back into the urn and then the second ball is taken out. So, the possibilities can be:

1st ball is black, 2nd ball is black: (B, B)

1st ball is black, 2nd ball is white: (B, W)

1st ball is white, 2nd ball is black: (W, B)

1st ball is white, 2nd ball is white: (W, W)

So, the sample space is: (B,B), (B,W), (W,B), (W,W).

No. of repetitions of the experiment in part (a) is 1. i.e. (B,B)

No. of repetitions of the experiment in part (b) is 2 i.e. (B,B) and (W,W).

The outcome of the first draw can be determined based on the probabilities for each ball.

P(Black) = No. of black balls/Total no. of balls

P(Black)= 2/3

P(White) = No. of white balls/Total no. of balls

P(White) = 1/3

Outcome of the first draw is most likely to be a black ball in either of these experiments because it has a higher probability.

(c) We can calculate the relative frequency of (W,W) as:

(W,W) = 1/3 x 1/3 = 1/9

(d) Yes, the outcome of the second draw depends on the first draw in the experiment in part (a) because the balls are drawn without replacement. If the first ball is white then the second ball can not be white hence, the outcome of the second draw depends on the first one.