Probability word problem?

Question

Ok, I need some help with this one. It's one of the ones on my homework that I just can't get past. I have done some brainstorming, but before we get onto that, here is the problem statement:

A friend recently planned a camping trip. He has two flashlights, one that requires a single 6V battery and another that used two D size batteries. He had previously packed two 6V and four D size batteries in his camper. Suppose the probability that any particular battery works is P and that all batteries work or fail independently of one another. Our friend wants to take just one flashlight. For what values of P should he take the 6V flashlight?

Here's what I have so far:

P(6V works) = 4 to choose, 2 to work = F1(P)
P(D size works) = 2 to choose, 1 to work = F2(P)

The wise decision for our friend would be to take 6V if F2(P) > F1(P).

Now I'm stuck. How should I proceed with this problem?

Answer 1

The probability for both 6V batteries to fail is (1-P)², so the odds that at least one will work is 1-(1-P)². The probability that exactly 3 D batteries will fail is 4P(1-P)³ and all 4 D batteries to fail is (1-P)^4, so that the probability that at least 2 will work is 1-4P(1-P)³-(1-P)^4. So, to find P, we equate the two, and solve:

1-(1-P)² = 1-4P(1-P)³-(1-P)^4

and get P = 2/3. Checking with a calculator shows that for P < 2/3, going with the 6V flashlight is safer. For P > 2/3, there is a very slim advantage to taking the D flashlight.

The reason why the probability for exactly 3 D batteries to work being
4P(1-P)³ is because there are 4 possible outcomes where the one D battery does work.

Also, consider the easy case where P = 1/2. Then the odds of at least one 6V battery working is 3/4, but the odds of at least 2 D batteries working is 11/16, just under 3/4.

Answer 2

Ok - here's how it goes:

The probability that at least 1 of the 6V will work is actually:

1 minus the probability that neither will work =

1 - (1-P)^2 = 1-1+2P-P^2 = P(2-P)

Probability that at least 2 Ds will work:

1 - minus (prob none work + prob EXACTLY 1 works) =

1 - ((1-P)^4+4P*(1-P)^3) =

= 1 - (1 - 4P + 6P^2 - 4P^3 + P^4 + 4P - 12P^2 + 12P^3 - 4P^4) =

= 3P^4 - 8P^3 +6P^2 = P^2(3P^2 -8P + 6)

So when is P(2-P) > P^2(3P^2 - 8P + 6) ?

2 - P > 3P^3 - 8P^2 + 6P

3P^3 - 8P^2 +7P -2 <0

You can take it from there.