In a company there are 7 executives: 4 women and 3 men. 3 are selected to attend a management seminar. Find these probabilities.
A) All 3 selected are men
B) all 3 selected are women
C) 2 men and 1 woman will be selected.
D) 1 man and 2 woman will be selected
Are they selected randomly?
If randomly, which would be an unusual way of selecting folks for training (think about that)..
a) 3/7 * 2/6 * 1/5
c) you can select those several ways..
m, m, w
m,w,m
w,m,m
so, probab=3/7*2/6*4/5 + 3/7*4/6*2/5 + 4/7*3/6*2/5 = amazing..3ways*4*3*2/7*6*5
You probably have some combination formulas for this in your text.
I do these by using combinations (since order does not matter)
e.g. #1
prob = C(3,3)/C(7,3) = 1/35
b) C(4,3)/C(7,3) = 4/35
c) C(3,2)*C(4,1)/C(7,3) = 3*4/35 = 12/35
d) C(3,1)*C(4,2)/C(7,3) = 3*6/35 = 18/35
jihi
A) The probability of all 3 selected being men can be calculated by dividing the number of ways to select 3 men out of the 3 available by the total number of ways to select any 3 executives. So the probability is:
(3 men selected) / (3 executives selected)
Since there are only 3 men available, the only possible outcome is selecting all 3 men, so the probability is 1.
B) Similarly, the probability of all 3 selected being women can be calculated by dividing the number of ways to select 3 women out of the 4 available by the total number of ways to select any 3 executives. So the probability is:
(3 women selected) / (3 executives selected)
Since there are only 4 women available, but we need to select 3, the probability would be 0. It's like trying to put 3 clowns into a tiny car, you can't fit them all!
C) The probability of 2 men and 1 woman being selected can be calculated by dividing the number of ways to select 2 men out of the 3 available and 1 woman out of the 4 available by the total number of ways to select any 3 executives. So the probability is:
(2 men selected) * (1 woman selected) / (3 executives selected)
This can be calculated as (3 choose 2) * (4 choose 1) / (7 choose 3) = (3*4) / (35) = 12/35, which can also be written as approximately 0.34. So the probability is approximately 34%.
D) The probability of 1 man and 2 women being selected can be calculated similarly to the previous case. So the probability is:
(1 man selected) * (2 women selected) / (3 executives selected)
This can be calculated as (3 choose 1) * (4 choose 2) / (7 choose 3) = (3*6) / (35) = 18/35, which can also be written as approximately 0.51. So the probability is approximately 51%.
Keep in mind that these probabilities assume that each executive has an equal chance of being selected and that the selections are made randomly. But let's be honest, life is full of surprises, just like a clown popping out of a box when you least expect it!
To find the probabilities of each scenario, we need to consider the total number of possible outcomes and the number of favorable outcomes. In this case, we're selecting 3 executives out of 7.
A) All 3 selected are men:
There are 3 men in total, so the number of favorable outcomes is 3. The total number of possible outcomes is the number of ways to choose 3 executives out of 7, which can be calculated using combinations. The formula for combinations is nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items being selected.
So, the probability of selecting all 3 men is:
P(A) = (number of favorable outcomes) / (total number of possible outcomes)
P(A) = (3C3) / (7C3) = 1 / (7C3) = 1 / 35
B) All 3 selected are women:
Similarly, there are 4 women in total, so the number of favorable outcomes is 4. The probability of selecting all 3 women is:
P(B) = (4C3) / (7C3) = 4 / (7C3) = 4 / 35
C) 2 men and 1 woman will be selected:
The number of favorable outcomes is the number of ways to choose 2 men out of 3 and 1 woman out of 4. So, the probability of selecting 2 men and 1 woman is:
P(C) = (3C2) * (4C1) / (7C3) = 3 * 4 / (7C3) = 12 / 35
D) 1 man and 2 women will be selected:
The number of favorable outcomes is the number of ways to choose 1 man out of 3 and 2 women out of 4. The probability of selecting 1 man and 2 women is:
P(D) = (3C1) * (4C2) / (7C3) = 3 * 6 / (7C3) = 18 / 35
Therefore, the probabilities are:
A) 1/35
B) 4/35
C) 12/35
D) 18/35