Answer to Question #206207 in Statistics and Probability for sss

Question

Question 1

In Uni, the probability that a student takes calculus or is on scholarship is 0.85. The probability that a student is on scholarship is 0.61 and the probability that a student is taking calculus is 0.31.If a student is randomly chosen, find the probability that the student is taking calculus and is on scholarship?

Question 2

A group of 9 people is going to be formed into committees of 4, 3, and 2 people. How many committees can be formed if no person can serve on more than one committee?

Answer 1

Question 1

Let $C$ denote the event that a student takes calculus.

Let $S$ denote the event that a student is on scholarship is 0.61.

Given

$$P(C\cup S)=0.85, P(C)=0.31,P(S)=0.61$$

Then

$$P(C\cap S)=P(C)+P(S)-P(C\cup S)$$

$$=0.31+0.61-0.85=0.07$$

The probability that the student is taking calculus and is on scholarship is $0.07.$

Question 2

Number of ways to choose 4 members from 9 is $\dbinom{9}{4}=\dfrac{9!}{4!(9-4)!}=126$

Number of ways to choose 3 members from the remaining 5 is $\dbinom{5}{3}=\dfrac{5!}{3!(5-3)!}=10$

Number of ways to choose 2 members from the remaining 2 is $\dbinom{2}{2}=1$

Total number (by multiplication rule) is

$$\dbinom{9}{4}\dbinom{5}{3}\dbinom{2}{2}=126(1)(1)=1260$$

 If no person can serve on more than one committee, then 1260 committees can be formed.