In a company, there are 7 executives: 4 women and 3 men. There are selected to attend a management seminar. Find the probabilities of
a) All men to be selected
b) 2 men and 1 woman will be selected
c) At least a man will be selected
d) At most 2 women will be selected
How many are you selecting?
From the choices I would conclude you are choosing 3 to form the group
a) prob(3 men) = C(3,3)*C(4,0)/C(7,3)
= 1/35
or Prob(3 m3n) = 3/7*2/6*1/5 = 6/210 = 1/35
b) Prob(2 men, 1 woman) = C(3,2)*C(4,1)/C(7,3)
= 3*4/35 = 12/35
or, could be MMW, MWM, or WMM
= 3/7 * 2/6 * 4/5 + 3/7 * 4/6 * 3/5 + 4/7 * 3/6 * 2/5)
= 4/35 + 4/35 + 4/35
= 12/35
c) exclude the case of all women
= 1 - C(4,3)/C(7,3)
=
d) let me know how you reasoned this one, and what you got
To find the probabilities for each scenario, we need to first determine the total number of possible outcomes and the number of favorable outcomes for each case.
Total Number of Outcomes:
Since there are 7 executives, the total number of possible outcomes is given by the number of ways to select a group of executives from a total of 7. This can be calculated using the combination formula:
nCr = n! / (r!(n-r)!)
where n is the total number of executives and r is the number of executives to be selected.
Let's calculate the total number of possible outcomes for each scenario:
a) All men to be selected:
In this case, we need to select all 3 men from the total of 3 men. So the number of possible outcomes is:
3C3 = 3! / (3!(3-3)!) = 1
b) 2 men and 1 woman will be selected:
We need to select 2 men from a total of 3 men and 1 woman from a total of 4 women. So the number of possible outcomes is:
3C2 * 4C1 = (3! / (2!(3-2)!)) * (4! / (1!(4-1)!)) = 3 * 4 = 12
c) At least a man will be selected:
In this case, we need to consider the outcomes where one or more men are selected. The number of outcomes where at least one man is selected is equal to the total number of outcomes minus the number of outcomes where no men are selected.
Total Number of outcomes = 2^7 (as each executive has 2 possible outcomes - selected or not selected)
Number of Outcomes where no men are selected = 4C3 * 3C0 = (4! / (3!(4-3)!)) * (3! / (0!(3-0)!)) = 4 * 1 = 4
Therefore, the number of outcomes where at least one man is selected is:
2^7 - 4 = 128 - 4 = 124
d) At most 2 women will be selected:
In this case, we need to consider the outcomes where 0, 1, or 2 women are selected. The number of outcomes where at most 2 women are selected is equal to the sum of the number of outcomes for each case.
Number of Outcomes where no women are selected = 3C3 * 4C0 = (3! / (3!(3-3)!)) * (4! / (0!(4-0)!)) = 1 * 1 = 1
Number of Outcomes where only 1 woman is selected = 3C2 * 4C1 = (3! / (2!(3-2)!)) * (4! / (1!(4-1)!)) = 3 * 4 = 12
Number of Outcomes where exactly 2 women are selected = 3C1 * 4C2 = (3! / (1!(3-1)!)) * (4! / (2!(4-2)!)) = 3 * 6 = 18
Therefore, the total number of outcomes where at most 2 women are selected is:
1 + 12 + 18 = 31
Now we can calculate the probabilities for each scenario by dividing the number of favorable outcomes by the total number of possible outcomes:
a) Probability of all men to be selected: 1/1 = 1
b) Probability of 2 men and 1 woman to be selected: 12/84 = 1/7 ≈ 0.1429
c) Probability of at least a man to be selected: 124/128 ≈ 0.9688
d) Probability of at most 2 women to be selected: 31/128 ≈ 0.2422