The University wishes to hire 4 faculty members from a group of 8 Male and 7 Female applicants i.e. total 8 + 7 = 15 applicants.
This can be done in "15 \\choose4" = 1365 number of ways, which is the number of total outcomes.
We have, the total number of ways of choosing a Men and b Women from a group of A Men and B Women is: "{A\\choose a}{B\\choose b}"
a) All four will be Men
Here, A = 8, B = 7, a = 4, b = 0
"\\therefore" The number of ways of choosing 4 Men = "{8\\choose 4}{7\\choose 0}" = 70
"\\therefore" The probability that all four will be Men = (no of favorable cases)/(no of total outcomes)
= 70/1365 = 0.0513 (rounded to 4 decimal places)
Answer: The probability that all four will be Men is 0.0513.
b) All four will be Women
Here, A = 8, B = 7, a = 0, b = 4
"\\therefore" The number of ways of choosing 4 Women = "{8\\choose 0}{7\\choose 4}" = 35
"\\therefore" The probability that all four will be Women = (no of favorable cases)/(no of total outcomes)
= 35/1365 = 0.0256 (rounded to 4 decimal places)
Answer: The probability that all four will be Women is 0.0256.
c) At least one will be a Women
The probability that at least one will be a Women
= 1 - The probability that there will be no Women [since, P(Ac) = 1 - P(A)]
= 1 - The probability that all four will be Men
= 1 - 0.0513 [by using answer a)]
= 0.9487
Answer: The probability that at least one will be a Women is 0.9487.
d) At least one will be a Men
The probability that at least one will be a Men
= 1 - The probability that there will be no Men [since, P(Ac) = 1 - P(A)]
= 1 - The probability that all four will be Women
= 1 - 0.0256 [by using answer b)]
= 0.9744
Answer: The probability that at least one will be a Men is 0.9744.
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