Question

Suppose that a class contains 15 boys and 30 girls, and that 10 students are to be selected at random for a special assignment. Find the probability that exactly 3 boys will be selected.

I need explanation and calculation, the calculation and explanation should equal atleast 300 words, nothing less and nothing irrelevant please, otherwise i will give thumbs down

Answer 1

Combination rule is:
ⁿC_r = n!/(n-r)!r!

Number of combinations to choose 10 students out of 45 students will be:

⁴⁵C₁₀ = 45!/((45 – 10)!(10!)) = 3190187286

Now, number of ways to choose 3 boys out of 15

¹⁵C₃ = 15!/((15-3)!(3!)) = 455

And,  the number of ways to choose 7 girls out of 30:

³⁰C₇ = 30!/((30-7)!(7!)) = 2035800

Probability = No. of favorable outcomes / No. of possible outcomes

=(No. of ways of selecting 7 girls AND 3 boys) / No. of ways of selecting 10 students out of 45 students

=(No. of ways of selecting 7 girls out of 30 * No. of ways of selecting 3 boys out of 15) / No. of ways of selecting 10 students out of 45 students

=(2035800*455)/3190187286 = 0.29

Probability of exactly 3 boys being selected = 0.29