an employer wishes to hire three people from a group of 15 applicants. 8 men and 7 women, all qualified to fill the position. if he selects the three at random. what is the probability. a) all will be men? b) one will be a women?
(a) p = 8/15 * 7/14 * 6/13
This is like choosing items without replacement, right?
or ..
prob(3 are all men) = C(8,3) / C(15,3)
= 56/455
= 8/65 , the same as Steve's
b) prob(one woman, two men)
= C(7,1)*C(8,2)/C(15,3)
= 7*28/455
= 28/65
a) Ah, the age-old quest for the perfect men's club! Let's calculate the probability of this happening randomly. We have 8 men out of 15 applicants, so the chance of selecting a man is 8/15 for the first pick. For the second pick, with one less man and one less applicant, the probability is 7/14. Finally, for the third pick, it becomes 6/13. To get the probability of all three being men, we multiply these probabilities together: (8/15) * (7/14) * (6/13) = 168/455. So, there's a 168/455 chance that all three will be men. Don't forget to balance it out with some female perspectives though!
b) Ah, the employer wants to add a pinch of femininity! To find the probability of just one woman being selected, we need to consider the different combinations. There are 7 women out of 15 applicants, so the probability of selecting a woman is 7/15 for the first pick. Then, since we only want one woman out of three picks, the probability of selecting two men consecutively is 8/14 * 7/13. Finally, no more women for the last pick, so the probability is 8/12. Multiplying these probabilities together gives us (7/15) * (8/14) * (7/13) * (8/12) = 1960/5460. Thus, there's a 1960/5460 chance that exactly one woman will be hired. Balance is key!
To find the probability for each scenario, we need to first determine the total number of possible outcomes. In this case, the employer needs to select 3 people from a group of 15 applicants. Therefore, the total number of possible outcomes can be calculated using the combination formula:
nCr = n! / (r!(n-r)!)
where n is the total number of applicants (15) and r is the number of people being selected (3).
a) Probability that all 3 will be men:
We have 8 men in the group of applicants, so the number of ways to select 3 men from 8 is given by the combination formula as follows:
8C3 = 8! / (3!(8-3)!) = (8 * 7 * 6) / (3 * 2 * 1) = 56
The total number of possible outcomes is:
15C3 = 15! / (3!(15-3)!) = (15 * 14 * 13) / (3 * 2 * 1) = 455
Therefore, the probability of selecting 3 men from the group is:
Probability (all men) = 56 / 455 ≈ 0.123
b) Probability that one will be a woman:
We can calculate this probability by considering the different cases where one woman is selected. We can select 1 woman from 7 and 2 men from 8, or 2 women from 7 and 1 man from 8. We will calculate the probability for each case and add them together.
Case 1: Selecting 1 woman and 2 men:
Number of ways to select 1 woman from 7: 7C1 = 7
Number of ways to select 2 men from 8: 8C2 = (8 * 7) / (2 * 1) = 28
Total number of outcomes for this case: 7 * 28 = 196
Case 2: Selecting 2 women and 1 man:
Number of ways to select 2 women from 7: 7C2 = (7 * 6) / (2 * 1) = 21
Number of ways to select 1 man from 8: 8C1 = 8
Total number of outcomes for this case: 21 * 8 = 168
The sum of the outcomes for both cases is: 196 + 168 = 364
Therefore, the probability of selecting one woman is:
Probability (one woman) = 364 / 455 ≈ 0.8
In summary:
a) The probability that all three selected will be men is approximately 0.123, or 12.3%.
b) The probability that one person selected will be a woman is approximately 0.8, or 80%.
To calculate the probability of each scenario, we first need to determine the total number of possible outcomes.
The total number of ways to choose 3 people out of 15 applicants is given by the binomial coefficient, which can be calculated using the combination formula:
C(15, 3) = 15! / (3! * (15-3)!) = 15! / (3! * 12!) = (15 * 14 * 13) / (3 * 2 * 1) = 455
a) Probability that all three selected will be men:
There are 8 men in the group of applicants, so the number of ways to choose 3 men out of 8 is given by:
C(8, 3) = 8! / (3! * (8-3)!) = 8! / (3! * 5!) = (8 * 7 * 6) / (3 * 2 * 1) = 56
Therefore, the probability of selecting all men is:
P(all men) = C(8, 3) / C(15, 3) = 56 / 455 ≈ 0.123
b) Probability that one selected person will be a woman:
There are 7 women in the group of applicants, so the number of ways to choose 1 woman out of 7 is given by:
C(7, 1) = 7! / (1! * (7-1)!) = 7! / (1! * 6!) = (7 * 6) / (1 * 1) = 42
Since we need to choose 2 people from the remaining 14 (excluding the chosen woman):
C(14, 2) = 14! / (2! * (14-2)!) = 14! / (2! * 12!) = (14 * 13) / (2 * 1) = 91
Therefore, the probability of selecting one woman is:
P(one woman) = (C(7, 1) * C(14, 2)) / C(15, 3) = (42 * 91) / 455 ≈ 0.364
Note: The probabilities are approximate decimal values. To make them more meaningful or easier to interpret, you can express them in percentage form by multiplying by 100%.