There are two flowers, labelled daisy and rose. There is a bird, initially in da
ID: 3013309 • Letter: T
Question
There are two flowers, labelled daisy and rose. There is a bird, initially in daisy, hunting a fly that is initially in the rose. They move from flower to flower independently: every minute each changes flowers (with probability p for the bird and q for the fly) or stays put, with the complementary probabilities. Once in the same flower, the bird eats the fly and the hunt ceases. (You can assume p=1/4, q=1/2)
A. What is the probability that the hunt lasts more than two minutes?
B. What is the probability that the ends within 3 minutes?
Explanation / Answer
SOLUTION
Back-up Theory
Hunt occurs in a particular minute, only if either bird changes flower and fly does not or vice versa. So, P(hunt in a particular minute) = p(1 - q) + q(1 - p) = p + q – 2pq = ¼ + ½ - 2(¼ x ½) = ½. And hence P(no hunt in a particular minute) = 1 - P(hunt in a particular minute) = 1- ½ = ½. These probabilities remain the same for all minutes.
Now,
Part A
Hunt lasts more than two minutes => No hunt in the first two minutes.
So, P(Hunt lasts more than two minutes) = P(No hunt in the first two minutes.) = P(No hunt in the first minute) x P(No hunt in the second minute) [because the given condition of independence] = ½ x ½ = ¼ ANSWER
Part B
The hunt ends within 3 minutes => hunt in first minute or no hunt in first minute but hunt in second minute or no hunt in first and second minute but hunt in third minute.
Hence, then required probability = ½ + (½ x ½) + (½ x ½ x ½) = ½ + ¼ + 1/8 = 7/8 ANSWER