Question

Q1 (10 points) 1. Butterflies. Alice, Bob, and Charlotte are looking for butterflies. They look in three separate parts of a field, so that their probabilities of success do not affect each other. Alice finds 1 butterfly with probability 17%, and otherwise does not find one. Bob finds 1 butterfly with probability 25%, and otherwise does not find one. Charlotte finds 1 butterfly with probability 45%, and otherwise does not find one. Let X be the number of butterflies that they catch altogether. Let Y be the number of people who do not catch a butterfly. Find the joint mass px,y(x, y) of X and Y.

Answer 1

Here let say Alice is A, Bob is B and Charlotte is C.

p(A) = 0.17; A = 1

= 0.83 ; A = 0

p(B) = 0.25; B = 1

= 0.75; B = 0

p(C) = 0.45; C = 1

= 0.55; C = 0

Here X = A + B + C

Y is the number of people don't catch butterfly.

so here as we can see that

X + Y = 3

so acceptable combinations are (0, 3), (1, 2), (2, 1) , (3, 0)

p(0, 3) = P(When none of them catch a butterfly) = 0.83 * 0.75 * 0.55 = 0.3424

p(1,2) = P(When one of them catch butterfly) = P(A catch butterfly rest don't) + P(B catch butterfly rest don't) + P(C catch butterfly rest don't)

= 0.17 * 0.75 * 0.55 + 0.83 * 0.25 * 0.55 + 0.83 * 0.75 * 0.45 = 0.4644

p(2,1) = P(When two of them catchbutterfly) = P(A and Bcatch butterfly rest don't) + P(B and C catch butterfly rest don't) + P(C and A catch butterfly rest don't)

= 0.17 * 0.25 * 0.55 + 0.83 * 0.25 * 0.45 + 0.17 * 0.75 * 0.45 = 0.1741

p(3,3) = P(All will catch butterfly) = 0.17 * 0.25 * 0.45 = 0.0191

so here

p(x,y) = 0.3424 ; x = 0, y = 3

= 0.4644 ; x = 1, y = 2

= 0.1741 ; x = 2, y = 1

= 0.0191 ; x = 3, y = 0