A history class is comprised of 5 female and 4 male students. If the instructor of the class randomly chooses 7 students from the class for an oral exam, what is the probability that 4 female students and 3 male students will be selected? Round your answer to 3 decimal places.(If necessary, consult a list of formulas.)
Question
A history class is comprised of 5 female and 4 male students. If the instructor of the class randomly chooses 7 students from the class for an oral exam, what is the probability that 4 female students and 3 male students will be selected? Round your answer to 3 decimal places.(If necessary, consult a list of formulas.)
Solution
To solve this problem, we need to use the combination formula which is C(n, r) = n! / [r!(n-r)!], where n is the total number of options, r is the number of options chosen, and "!" denotes a factorial.
First, we calculate the total number of ways to choose 7 students out of 9, regardless of gender. This is C(9, 7) = 9! / [7!(9-7)!] = 36.
Next, we calculate the number of ways to choose 4 female students out of 5, which is C(5, 4) = 5! / [4!(5-4)!] = 5.
Then, we calculate the number of ways to choose 3 male students out of 4, which is C(4, 3) = 4! / [3!(4-3)!] = 4.
The number of favorable outcomes (choosing 4 female and 3 male students) is the product of these two, which is 5 * 4 = 20.
Finally, we divide the number of favorable outcomes by the total number of outcomes to get the probability. So, the probability is 20 / 36 = 0.555.
Rounding to three decimal places, the probability is 0.556.
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