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9. The shell of the snail Cepaea nemoralis can be yellow, pink, or brown. Suppos

ID: 3041071 • Letter: 9

Question

9. The shell of the snail Cepaea nemoralis can be yellow, pink, or brown. Suppose that a very large natural population of this species consists of 27% yellow, 39% pink, and 34% brown individuals. If you randomly sample 2 snails from this population, what is the probability that (a) both are pink? (b) neither is pink? (c) one is pink and one is not? [Hint: when taking a small sample from a very large population, sampling without replacement is essentially the same as sampling with replacement since removing a few individuals does not significantly alter the makeup of the population.]

10. The probability that a train from Paris to Berlin will leave on time is 0.62, the probability that it will arrive on time is 0.84, and the probability that it will both leave on time and arrive on time is 0.60. (a) What is the conditional probability that if such a train leaves on time it will also arrive on time? (b) What is the conditional probability that if such a train does not leave on time it will nevertheless arrive on time? (c) What is the conditional probability that if such a train arrives on time it also left on time?

Explanation / Answer

9)a)probability that both are pink =0.39*0.39=0.1521

b)neither is pink =(1-0.39)*(1-0.39)=0.3721

c)one is pink and one is not =P(fist is pink and second not+first not and second is)

=0.39*(1-0.39)+(1-0.39)*0.39 =0.4758

10)

let probability of arrive on time is P(A) and leave on time =P(L)

here P(A) =0.84 and P(L) =0.62

P(A n L) =0.6

a) conditional probability that if such a train leaves on time it will also arrive on time

=P(A|L) =P(AnL)/P(L) =0.6/0.62=0.9677

b)  conditional probability that if such a train does not leave on time it will nevertheless arrive on time

=P(A|Lc) =P(AnLc)/(1-P(L)) =(P(A)-P(AnL))/(1-P(L))=(0.84-0.60)/(1-0.62)=0.6316

c) conditional probability that if such a train arrives on time it also left on time

=P(L|A) =P(L n A)/P(A) =0.6/0.84 =0.7143