   Chapter 8, Problem 2RE ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# A bag contains 5 black balls and 7 white balls. If we draw 4 balls, with each one replaced before the next is drawn, what is the probability that(a) 2 balls are black?(b) at least 2 balls are black?

(a)

To determine

To calculate: The probability of drawing 2 black balls in 4 trials out of a bag, where a bag contains 5 black balls and 7 white balls.

Explanation

Given Information:

It is given that a bag contains 5 black balls and 7 white balls, also someone draw 4 balls with each one replaced before the next is drawn.

Formula used:

For a binomial distribution the probability is,

Pr(x)=(Cxn)pxqnx

Where n the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure.

q=1p

And,

Cxn=nxnx

Calculation:

Consider A bag contains 5 black balls and 7 white balls. If we draw 4 balls with each one replaced before the next is drawn.

Then the probability of drawing of a black ball out of total 12 balls when each time after drawing the ball is replaced is p=512.

since there are total 5 black balls and there will always be 5 black balls only because of replacement.

Then the probability of failure i.e. of drawing white ball is,

q=1p=1512=712

Now, the probability of drawing 2 black balls in 4 trials out of a bag,

Pr(x)=(Cxn)pxqnx

Where n=4,x=2,p=512,q=712

So,

Pr(2)=(C24)(

(b)

To determine

To calculate: The probability of drawing at least 2 black balls in 4 trials out of a bag, where A bag contains 5 black balls and 7 white balls.

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