probability question

Question

4 different written driving tests are administered by the Motor Vehicles Division.  One of these 4 tests is selected at random for each applicant for a driver's license.  If a group consisting of two women and three men apply for license, what is the probability that:

a.  Exactly two of the five will take the same test? (show all work)

b.  The two women will take the same test. (show all work)

 

 

 

 

There are 10 batteries in a box and two of them are defective.  If a random sample of 3 batteries is drawn from the box (without replacement), and X represents the number of defective batteries in the sample:

a. What is the probability distribution of X?

b. What is the expected value of X?

c. What is the variance of X?

Answer 1

In the first question, let's number each person 1 - 5.  The probability that person 1 and person 2 will take the same test is 1/4 (once person 1's test is known, there's a 1/4 chance that person 2 will get the same one).  Now that person 1 and person 2 have the same test, each other person must receive a different test.  So, the probability that person 3 will get a different test is 3/4, the probability that person 4 will get a different test is 2/4, and finally, the probability that person 5 will get a different test still is 1/4.  Multiply these all together and you get 6/256.  This represents the probability that person 1 and person 2 will receive the same test, with everyone else receiving a different test.

 

However, it doesn't have to be person 1 and person 2 who have the same test.  It could any two of the five that receive the same test.  The number of combinations of 2 people getting the same test you can make out of the 5 applicants is 5 choose 2 [5!/(2!*3!)], which is 10.  So we need to multiply 6/256 by 10, which is 60/256, which is about 0.23.

 

 

In part b of the first question, we can ignore what test(s) the men receive.  Suppose one of the women receives a certain test.  The probability that the second woman receives the same test is 1/4.

 

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In the second question, the distribution is hypergeometric because there is no replacement.  N is the population size (10), n is the number of batteries in the population that are defective (2), r is the sample size (3), and x is the number of defective batteries in the sample.

 

The distribution is:

 

(n choose x)(N - n choose r - x)/(N choose r)

 

This distribution has an intuitive structure.  The numerator represents how many ways you can "succeed" in this experiment (picking exactly x defective batteries), and the denominator represent the total number of outcomes in this experiment.

 

(n choose x) represents the number of ways you could select x defective batteries from the n total defective batteries in the population.

 

 

With x defectives in the sample, there are r - x non-defectives in the sample, and with n defectives in the population, there are N - n non-defectives in the population.  Thus, (N - n choose r - x) is how many different ways you could pick the non-defective batteries in your sample from the entire population of non-defective batteries.

 

(N choose r) is the number of ways you can select a sample of r batteries from a population of N.

 

Plug in your numbers to get (2 choose x)(8 choose 3-x)/(10 choose 3), which is your distribution.

 

Part b, E(X) = nr/N = (2)(3)/10 = 3/5

Part c, Var(X) = (nr/N)((N-r)/N)((N-n)/(N-1)) = (2*3/10)((10-3)/10)((10-2)/(10-1)) = (6/10)(7/10)(8/9) = 336/900 = 0.373