A sixth form class consists of 6 girls and 9 boys. Three students from the class are chosen at random. The number of boys chosen is denoted by the random variable X. Show that
a)P(X=0)=120/2730
b)P(X=2)=1296/2730
To solve this problem, we will use combinations.
First, let's calculate the total number of ways to choose 3 students out of 15 (6 girls + 9 boys):
Total number of ways = C(15, 3)
Now, let's calculate the number of ways to choose 3 students such that X = 0 (no boys are chosen).
Number of ways to choose 3 students without any boys= C(6, 3)
Since there are 6 girls, we choose all 3 of them.
Now, let's calculate the number of ways to choose 3 students such that X = 2 (2 boys are chosen).
Number of ways to choose 2 boys from 9= C(9, 2)
Number of ways to choose 1 girl from 6= C(6, 1)
Now, let's calculate the probabilities:
a) P(X = 0) = (Number of ways to choose 3 students without any boys) / (Total number of ways)
= C(6, 3) / C(15, 3)
b) P(X = 2) = (Number of ways to choose 2 boys and 1 girl) / (Total number of ways)
= (C(9, 2) * C(6, 1)) / C(15, 3)
Now let's calculate these probabilities:
a) P(X = 0) = (C(6, 3)) / (C(15, 3))
= (6! / (3! * (6 - 3)!)) / (15! / (3! * (15 - 3)!))
= (6 * 5 * 4 / (3 * 2 * 1)) / (15 * 14 * 13 / (3 * 2 * 1))
= 20 / 455
= 4 / 91
≈ 0.043956
We can simplify 4/91 by dividing both numerator and denominator by 4 to get 1/23.
Therefore, P(X = 0) ≈ 1/23 ≈ 0.043956.
b) P(X = 2) = ((C(9, 2) * C(6, 1)) / (C(15, 3))
= (9! / (2! * (9 - 2)!)) * (6! / (1! * (6 - 1)!)) / (15! / (3! * (15 - 3)!))
= (9 * 8 / (2 * 1)) * (6 / 1) / (15 * 14 * 13 / (3 * 2 * 1))
= (72 / 2) * 6 / (2730 / 6)
= (72 * 6 * 6) / 2730
= 1296 / 2730
= 48 / 101
≈ 0.475248
Therefore, P(X = 2) ≈ 48/101 ≈ 0.475248.
Hence, we have shown that
a) P(X = 0) = 120/2730
b) P(X = 2) ≈ 1296/2730.