Four different written driving tests are administered by the Motor Vehicle Department. One of these four tests is selected at random for each applicant for a driver's license.
If a group of 5 applicants (consisting of 2 women and 3 men) apply for a license, what is the probability that...
a) Exactly 2 of the 5 applicants take the same test?
b) The 2 women take the same test?
Count the number of ways you can assign test to the people and then divide by the total number of ways (which is 4^5).
There are 4 ways to choose which test is going to be done by two given applicants who take the same test. There are Binomial[5,2] ways to choose these two applicants.
The remaining 3 applicants must be assigned any one of the 3 other tests. There are 3^3 ways for this.
Now,
4*Binomial[5,2]*3^3
is not the correct answer, because of the double counting of the cases where 2 out of 3 of the remaining group of 3 are also assigned the same test.
Depending on how the question is to be interpreted, such cases should not be counted at all, or they should be counteed once. So, you need to compute the number of these cases and then subtract that once or twice from the above expression.
You can also try a more formal approach using the principle of inclusion and exclusion.
a 15/256, b 1/6
4 and5
To solve this problem, we need to calculate the probabilities for each case separately. Let's start by analyzing each scenario.
a) Exactly 2 of the 5 applicants take the same test:
To calculate this probability, we need to find the number of favorable outcomes and divide it by the total number of possible outcomes.
Favorable outcomes: We have two women and three men. We can choose the two women to take the same test in ${2 \choose 2} = 1$ way. Then, we need to distribute the other three men among the remaining tests. We can do this in ${3 \choose 1} = 3$ ways. So the number of favorable outcomes is $1 \times 3 = 3$.
Total number of possible outcomes: Each applicant can take any of the four tests, so every applicant has 4 possible tests. Therefore, the total number of possible outcomes is $4^5$ since there are five applicants in total.
Probability = Favorable outcomes / Total outcomes = 3 / 4^5 = 3 / 1024 ≈ 0.00293
Therefore, the probability that exactly 2 of the 5 applicants take the same test is approximately 0.00293.
b) The 2 women take the same test:
To calculate this probability, we need to find the number of favorable outcomes and divide it by the total number of possible outcomes.
Favorable outcomes: The two women need to take the same test, and the three men can take any of the remaining tests. We can choose the test for the women in ${4 \choose 1} = 4$ ways. Then, the three men can choose their tests from the remaining three options in $3^3$ ways. So the number of favorable outcomes is $4 \times 3^3 = 108$.
Total number of possible outcomes: Same as before, each applicant can take any of the four tests, so every applicant has 4 possible tests. Therefore, the total number of possible outcomes is $4^5$ since there are five applicants in total.
Probability = Favorable outcomes / Total outcomes = 108 / 4^5 = 108 / 1024 ≈ 0.10547
Therefore, the probability that the two women take the same test is approximately 0.10547.