Question

a history class is comprised of 7 female and 9 male students. if the instructor of the class randomly chooses 10 students from the class for an oral exam, what is the probability that 3 female students and 7 male students will be selected? round your answer to 3 decimal places.

(if necessary, consult a list of formulas.)

Explanation:

Identify the total number of outcomes

Using the Combinations knowledge point

$$ \text{Total students} = 7 + 9 = 16 $$

$$ N = \binom{16}{10} = \frac{16!}{10! \cdot 6!} = 8008 $$

Calculate the number of favorable outcomes

Using the Combinations knowledge point

$$ \text{Ways to choose 3 females} = \binom{7}{3} = \frac{7!}{3! \cdot 4!} = 35 $$

$$ \text{Ways to choose 7 males} = \binom{9}{7} = \frac{9!}{7! \cdot 2!} = 36 $$

$$ \text{Favorable outcomes} = 35 \cdot 36 = 1260 $$

Compute the probability

Using the Probability of Events knowledge point

$$ P = \frac{1260}{8008} \approx 0.15734 $$

Round to three decimal places

$$ P \approx 0.157 $$

Answer:

A history class is comprised of 7 female and 9 male students. If the instructor of the class randomly chooses 10 students from the class for an oral exam, what is the probability that 3 female students and 7 male students will be selected? Round your answer to 3 decimal places.

(If necessary, consult a list of formulas.)

<blank>0.157</blank>