Question
#3 please show work no calculator
Plesse write your solutions in the blhue books distributed to you. Yoa ean put your solutions in whatever onder you tike, but make sure you mumber your solutions so I know which is which. Show as much work an you can, unless otherwise speciied Caleulators and notes sre not permited Formula. If Bi… , B-partition the probability "pace, and 1. Exene talkes four basketball free-throws in a row. She is successful on each one independently with probability 4/5 You can (and in fact, should!) leave your answers unsimplified on this question. That is if the answer is 1 - (3) leave your answer in this form and don't attempt to simplify it. (a) What's the probability that she makes all of her shots? (b) What's the probability that she mabes at least one shot? (c) What's the probability that in order, ber sequence of shots is success, miss, miss, success? (d) What's the probability that she misses her last two shots? 2. If gene HNFIA is present in a mouse, then it has probability 9 of making it through a given maze. If not, the probability is only .6. In a mouse selected at random from the population, the gene is present with probability 2. What is the probability that the mouse can make it through the maze? 3. For each part, write down possible or impossible depending on whether the scenario could occur or not. You do not need to justify your answer: just write possible or impossible. (a) P(A) = 1/4,P(B) = 1/2, P(AUB) = 3/4 A and B are disjoint. (b) P(A)-1/4, P(B) = 1/2; if B occurs, then A also occurs. (c) P(A-1/4, P(B)-1/2, and P(AU B) = 1. (d) P(A) = 1/4, P(B) = 1,2, P(An B)-1/8; A and B are disjoint. (e) P(A) = 3/4,P(B) = 1/2; A and B are disjoint. (f) P(A) = 34, P(B)s 1/2, p(AUB) = 7/s A and B are independent. There are more problems on the back!
Explanation / Answer
3) As A and B are disjoint therefore P(AnB)=0
hence P(AUB)=P(A)+P(B) which is true for above hence possible
b) as if for B occurs ; A also occurs condition should be P(A)>=P(B). but from above P(A)<P(B) ; therefore impossible
c)for P(AUB) is always <= P(A)+P(B) hence impossible
d)for disjoint events P(AnB)=0 ; therefore impossible
e)for disjoint events P(AUB)=P(A)+P(B)=3/4+1/2=5/2 which is greater then 1. and as we know that probability can not be greater then 1; therefore impossible
f) from above P(AnB)=P(A)+P(B)-P(AnB)=3/4+1/2-7/8=3/8
as P(AnB)=P(A)*P(B) ' therefore A and B are independent.
hence possible
