Question

I have a box with 5 dog toys. Pugsley chooses randomly one of them. I put it back and then Bean chooses randomly one of them. I put it back and repeat the experiment with my third dog, Sprout. What's the probability that at least 2 of the three dogs have chose the same toy?

A.16/25

B.13/25

C.7/25

Now I open the door to the deck. Pugsley has a probability of .9 of going outside, Bean of .8, Sprout of .5. They choose to go or not independently. What's the probability that exactly 2 of them will go?

A.49

B.51

C.72

Answer 1

1. We have, 5 toys, so anyone same toy which can be chosen by the dogs= 1/5

Now, we have, n= 3 (3 toys chosen by 3 dogs)

Let X: number of times same toy is chosen

So, X~Binomial(3,1/5) distribution

Pdf of X: P(X=x) = 3Cx*(1/5)^x*(4/5)^3-x

We want, P(X>=2) = P(X=2) + P(X=3) = 3C2*(1/5)^2*(4/5) + 3C3*(1/5)^3 = 13/125

note: It's incorrectly written as /25 in the denominatior, it should be /125 in the options given.

2. We can have the following three scenarios:

Pugsley and Bean go out, Sprout does not

Pugsley and sprout go out, Bean does not

Bean and sprout go out, Pugsley does not

So the following probabilies added are: 0.9*0.8*0.5 + 0.9*0.5*0.2 + 0.8*0.5*0.1= 0.49

So option A is correct.