A company is hiring to fill four managerial vacancies. The candidates are five men and three women. If every candidate is equally likely to be chosen then the probability that at least one women will be selected is _______ (round off to 2 decimal places).

The correct answer is 0.93.

Step-by-Step Explanation:

1. Identify the total pool of candidates and selection requirements:
We are given a pool of candidates consisting of:
- Men = 5
- Women = 3
- Total candidates = 5 + 3 = 8
The company needs to fill 4 managerial vacancies, meaning we need to select exactly 4 candidates from the total of 8.

2. Calculate the total number of ways to select 4 candidates:
The total number of ways to choose 4 candidates out of 8 is given by the combination formula ⁿC_r:
$${}^8{C}_4=\frac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1}$$

$${}^8{C}_4=70$$
So, there are 70 total possible combinations of selecting 4 candidates.

3. Use the complement method to find the required probability:
To find the probability that at least one woman is selected, we can use the complement event:
Probability(at least 1 woman) = 1 - Probability(no woman is selected)

4. Calculate the probability of selecting no women (all men):
Selecting no women means that all 4 selected candidates must be men. Since there are 5 men available, the number of ways to choose 4 men is:
$${}^5{C}_4=5$$
Thus, the probability of selecting no women is:
$$\text{Probability(no woman)}=\frac{{}^5{C}_4\times {}^3{C}_0}{{}^8{C}_4}=\frac{5\times 1}{70}=\frac{5}{70}$$

5. Calculate the probability of selecting at least one woman:
Now, subtract this probability from 1:
$$\text{Probability(at least 1 woman)}=1-\frac{5}{70}=\frac{65}{70}\approx 0.92857$$
Rounding off to 2 decimal places, we get 0.93.