# Thread: neurotic dogs

1. ## neurotic dogs

Problem:
I have two dogs and they're very neurotic. Mingus isn't happy unless at least 2 of his 3 favorite chewtoys are in the backyard with him, and Duke isn't happy unless at least 3 of his 5 favorite chewtoys are in the frontyard with him.

The probability that any particular chewtoy is not in its respective dog's yard is 1-a. (Note that the probability is independent from chewtoy to chewtoy, but the probabilities themselves are all the same). For what values of a is Mingus' happiness more probable than Duke's happiness?

What I think:
Intuitively, I want to say that since the probabilities are all the same that it's always a safer bet on Mingus' happiness over Duke's, but I've learned that my intuition sucks at probability. So I tried calculating the probability that each dog will be happy. I'll use my Mingus calculation as an example:

Prob(Mingus happy) =
=1-Prob(Mingus not happy)
= 1-Prob(2 Mingus toys missing or 3 Mingus toys missing)
= 1-( a(1-a)^2 + (1-a)^3 ) = 2a-a^2 (if I did my algebra right)

I also tried:
Prob(Mingus happy)=
=Prob(2 Mingus toys in yard or 3 Mingus toys in yard)
=a^2 * (1-a) + a^3 = a^2.

So I have two issues, and any clarification on these would be awesome:
(1) I got something different when I tried to calculate the same thing two different ways, and

(2) by either way of doing, I set Prob(Mingus happy) > Prob(Duke happy) and tried to solve for a range of values for a but got either something unhelpful or an inequality that agreed with my earlier intuition.

2. Hello cribby

Welcome to Math Help Forum!
Originally Posted by cribby
Problem:
I have two dogs and they're very neurotic. Mingus isn't happy unless at least 2 of his 3 favorite chewtoys are in the backyard with him, and Duke isn't happy unless at least 3 of his 5 favorite chewtoys are in the frontyard with him.

The probability that any particular chewtoy is not in its respective dog's yard is 1-a. (Note that the probability is independent from chewtoy to chewtoy, but the probabilities themselves are all the same). For what values of a is Mingus' happiness more probable than Duke's happiness?

What I think:
Intuitively, I want to say that since the probabilities are all the same that it's always a safer bet on Mingus' happiness over Duke's, but I've learned that my intuition sucks at probability. So I tried calculating the probability that each dog will be happy. I'll use my Mingus calculation as an example:

Prob(Mingus happy) =
=1-Prob(Mingus not happy)
= 1-Prob(2 Mingus toys missing or 3 Mingus toys missing)
= 1-( a(1-a)^2 + (1-a)^3 ) = 2a-a^2 (if I did my algebra right)

I also tried:
Prob(Mingus happy)=
=Prob(2 Mingus toys in yard or 3 Mingus toys in yard)
=a^2 * (1-a) + a^3 = a^2.

So I have two issues, and any clarification on these would be awesome:
(1) I got something different when I tried to calculate the same thing two different ways, and

(2) by either way of doing, I set Prob(Mingus happy) > Prob(Duke happy) and tried to solve for a range of values for a but got either something unhelpful or an inequality that agreed with my earlier intuition.
Thanks for showing us your working. You're going wrong in the same way in both of your methods.

The probability that any given toy is in the yard is $\displaystyle a$. So the probability that $\displaystyle 2$ out of $\displaystyle 3$ toys are in the yard is $\displaystyle 3 a^2(1-a)$. You forgot to multiply by $\displaystyle 3$. The reason the $\displaystyle 3$ is there is because there are $\displaystyle 3$ ways in which this event can happen, depending upon exactly which of the $\displaystyle 3$ toys is not in the yard. So the probability that Mingus is happy is
$\displaystyle p(2$ out of $\displaystyle 3) + p(3$ out of $\displaystyle 3) = 3a^2(1-a)+a^3 = 3a^2 -2a^3$
In general if we want the probability that $\displaystyle r$ out of $\displaystyle n$ toys are in the yard we use the Binomial Coefficient $\displaystyle \binom{n}{r}$, and get the probability
$\displaystyle \binom{n}{r}a^r(1-a)^{n-r}$
So, for example, the probability that 3 out of Duke's 5 toys are in the yard is
$\displaystyle \binom53a^3(1-a)^2=10a^3(1-2a+a^2)$
So work out the probability that $\displaystyle 3, 4$ or $\displaystyle 5$ of Duke's toys are in the yard; add these together and set the result to be less than the result for Mingus that I've given you above.

When I do this and simplify the result I get the inequality
$\displaystyle 2a^3-5a^2+4a-1 <0$.
This gives the answer $\displaystyle 0<a<\tfrac12$.

Do you want to try again and see if you agree with my answer?

Grandad

3. Oh for the love of...

Yes, the binomial coefficient. If they had a prescription medication for absent-mindedness I'd be first in line.

I went back and worked through it correctly this time--much more agreeable to finagle now. Thanks so much for your time, and for the warm welcome to the forum as well!