SOLUTION: At Med U. the failure rate is 20%. Of 15 students registered in the freshman class; What is the probability that they all pass? What is the probability that a least 1 fails? Ho

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Question 450068: At Med U. the failure rate is 20%. Of 15 students registered in the freshman class;
What is the probability that they all pass?
What is the probability that a least 1 fails?
How many freshman are likely to pass?
An algebra class has 18 girls and 14 boys. Five tickets to a concert have just been made available to the class. The students to get the tickets will be chosen at random. What is the probability that:
a) five girls are chosen?
b) five boys are chosen?
c) at least 1 girl is chosen?
d) How many girls are likely to be chosen?
Thanks so much for your help!
Shauna

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
Note: The probability of x successes in n trials is:
P = nCx* where p and q are the probabilities of success and failure respectively.
In this case p(failing) = .20 and q(passing) = .80 |the failure rate is 20%.
nCx =
Of 15 students registered in the freshman class;
P(all pass) = (.80)^15 = .0352
P(at least 1 fails) = 1 - P(none failing) = 1 - .0352 = .9648
How many freshman are likely to pass? .80*15 = 12
An algebra class has 18 girls and 14 boys.
Five tickets to a concert have just been made available to the class.
The students to get the tickets will be chosen at random.
What is the probability that:
a) five girls are chosen = 18C5/32C5 = 8568/201,376 = .0425
nCx = 18C5 = 18!/(5!13!) and 32C5 = 32!/5!27!
These can be found with a calculator that has a factorial function.
b) five boys are chosen = 14C5/32C5 = 2002/201,376 = .0099
c) at least 1 girl is chosen? 1 - P(all boys) = 1 - .0099 = .9901
d) How many girls are likely to be chosen? (18/32)*5 = 2.8

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