Do Problem 22if one person is busy 3 evenings, one is busy2evenings, two are each busy one evening, and the rest are free every evening.

Short Answer

Expert verified

The required probability is 7202401.

Step by step solution

01

Given Information

One person is busy 3 evenings, one is busy 2 evenings, two are each busy one evening, and the rest are free every evening.

02

Definition of uniform sample spaces.

If a given experiment's sample space is known to be uniform, the probability of an event can be calculated using the event sizes and the sample space.

03

Find the Probability.

one person is busy 3 evenings, one is busy 2 evenings, two are each busy one evening, and the rest are free every evening.

The required probability is 4757672779=7202401.

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Most popular questions from this chapter

Let x1,x2,..,xnbe independent random variables, each with density function f(x), expected valueμ , and varianceσ2 . Define the sample meanby.x=i=1nxiShowthatE(x)=μ,and .var(x)=σ2n (See Problems 5.9,5.13and6.15.)

(a) Set up a sample space for the 5 black and 10 white balls in a box discussed above assuming the first ball is not replaced. Suggestions: Number the balls, say 1 to 5 for black and 6 to 15 for white. Then the sample points form an array something like (2.4), but the point 3,3 for example is not allowed. (Why?

What other points are not allowed?) You might find it helpful to write the

numbers for black balls and the numbers for white balls in different colors.

(b) Let A be the event “first ball is white” and B be the event “second ball is

black.” Circle the region of your sample space containing points favorable to

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number of sample points in A and in B; these are and . The region

AB is the region inside both A and B; the number of points in this region is

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and and verify (3.3) numerically.

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