1) The chances you go to the game on Saturday is 0.3, and the probability of staying home on Sunday is 0.4. Find the probability of doing both:
a. If they are mutually exclusive
b. if they are independent
For this problem, p(a) = game and p(b) = staying.
a. If they're mutually exclusive, is it just Ø?
b. P(AintersectionB) = P(a) * p (b) =
0.3 * 0.4 = 0.12
2) Suppose 30% of students attend more than one institution during their college career. A sample of 10 students is chosen. Assume that each student's college attendance pattern is independent of each other
a. Exactly 5 of the 10 attend more than one institution
b. At least two of the students attend more than 1 institution.
I think this is a binomial problem right?
n= 10 p=0.30 q=0.70
a) p(x=5) c(10,5) (.30)^5 (.70) ^10-5
so set up like this?
b) At least - so a trick I use to remember this to is to think of more. So At least two is p(x>2)?
I like to do it the quick way:
1- p (x=0) + p (x+1)
p(x=0 ) = c(10,0) (.30)^0 (.70)^10-0
p(x=1 ) = c(10,1) (.30)^1 (.70)^10-1
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
What are your answers?
I already wrote my answers?????
2) Yes, this is a binomial problem. We can use the binomial probability formula to solve it.
a) To find the probability that exactly 5 of the 10 students attend more than one institution, we can use the formula:
P(x = 5) = C(10, 5) * (0.30)^5 * (0.70)^(10-5)
The C(10, 5) represents the number of ways to choose 5 students out of 10, which is given by the combination formula. (0.30)^5 represents the probability that a student attends more than one institution (success) and (0.70)^(10-5) represents the probability that a student does not attend more than one institution (failure).
b) To find the probability that at least two of the students attend more than one institution, we can use the complement rule. The complement of "at least two" is "none or exactly one", so we can subtract the probability of "none or exactly one" from 1.
P(at least two) = 1 - [P(x = 0) + P(x = 1)]
To find P(x = 0), we use the formula:
P(x = 0) = C(10, 0) * (0.30)^0 * (0.70)^(10-0)
To find P(x = 1), we use the formula:
P(x = 1) = C(10, 1) * (0.30)^1 * (0.70)^(10-1)
Then we can substitute these values into the formula to get the final answer.