At least I think that is what I should do. We are working on probability and I am completely confused. We were given the following practice problem but I am lost most of the other problems gave more information. I have no idea which equation to use. I have looked at this problem for a while and all I get is that answer c is 20.4%? What am I missing?

Problem: A certain airplane has two independent alternators to provide electrical power. The probability that a given alternator will fail on a 1-hour flight is .02. What is the probability that

(a) both will fail? (b) Neither will fail? (c) One or the other will fail? Show all steps carefully.

Looks like a binomial to me. Two outcomes: it either fails or it does not fail. They are independent of one another.

(a) 2 of 2 fails, (b) 0 of 2 fails, (c) 1 of 2 fails

Might be another way to do it, but I definitely don't get the intersection/union thing. I don't see where you got 20.4%, but if the chances were that high that alternators are failing, I don't think I'd be flying.

Ok, let

A denote the event that the first alternator fails andb
B denote the event that the second alternator fails.

(a) The probability that both fail is represented by P(A \cap B).

(b) The probability that neither fails (both don't fail) is represented by P(\bar A \cap \bar B)

where \bar A and \bar B represent the event where the alternator does not fail and is equal to 1 - P(A) and 1 - P(B) respectively.

(c) The probability that one or the other fails is represented by P(A \cup B).

Whenever you have the word 'both', 'neither', 'and', you denote it by intersection and you multiply the sets (here, a set is the probability of an event)

Whenever you have the word 'or', you denote it by union, and you add the sets.

e.g..

P(X \cap Y) = P(X) \times P(Y)

P(X \cup Y) = P(X) + P(Y)

Ok, let

A denote the event that the first alternator fails andb
B denote the event that the second alternator fails.

(a) The probability that both fail is represented by P(A \cap B).

(b) The probability that neither fails (both don't fail) is represented by P(\bar A \cap \bar B)

where \bar A and \bar B represent the event where the alternator does not fail and is equal to 1 - P(A) and 1 - P(B) respectively.

(c) The probability that one or the other fails is represented by P(A \cup B).

Whenever you have the word 'both', 'neither', 'and', you denote it by intersection and you multiply the sets (here, a set is the probability of an event)

Whenever you have the word 'or', you denote it by union, and you add the sets.

eg.

P(X \cap Y) = P(X) \times P(Y)

P(X \cup Y) = P(X) + P(Y)

so if the events are independent and the probability a given alternator fails is .02 then P(A) = .02, P(B) = .02?

P(AintersectB) = P(A)xP(B) = (.02)(.02) = 0.0004

P(AintersectB w/bar)= P(AintersectB) = 1-0.0004 = 0.9996

P(AunionB) = P(A) + P(B) = .02 + .02 = 0.04

The probability that both alternators fail is .04%
The probability that neither alternator fails is 99.9%
The probability that one or the other fails is 4%

P(AintersectB w/bar)= P(AintersectB) = 1-0.0004 = 0.9996

This doesn't work. 1 - \cap is the probability that one or the other or neither. The intersection is BOTH. The opposite of "both" is not "neither." If one fails, that's not "both," right? It's also not "neither." This isn't what Unky said. He didn't write P\ (A\ \cap\ B)'. He wrote P\ (A'\ \cap\ B'). Those are two different things. What's the compliment of A? What's the compliment of B? Now what's the intersection of those?




P(AunionB) = P(A) + P(B) = .02 + .02 = 0.04

The only way you can do this is if there is no intersection, which there is. What's your addition equation?

Done this way, it's all the same answers I got using a binomial. (Done correctly, that is.) So there's obviously more than one way to solve it, but I would've never come up with this way.

Oh, shoo, yes, I forgot to say that part.

If you have a Venn Diagram in front of you, you'll see it more clearly.

When you have P(A \cup B), you must also remove one intersection. I'm really sorry for my mistake.

I myself when confronted to or, I think like this:

It's either:
A fails, and B works or
A works, but B fails.

Then, the probability becomes:
P(A or B) = P(A)P(\bar B) + P(\bar A)P(B)

This doesn't work. 1 - \cap is the probability that one or the other or neither. The intersection is BOTH. The opposite of "both" is not "neither." If one fails, that's not "both," right? It's also not "neither." This isn't what Unky said. He didn't write P\ (A\ \cap\ B)'. He wrote P\ (A'\ \cap\ B'). Those are two different things. What's the compliment of A? What's the compliment of B? Now what's the intersection of those?




The only way you can do this is if there is no intersection, which there is. What's your addition equation?

Done this way, it's all the same answers I got using a binomial. (Done correctly, that is.) So there's obviously more than one way to solve it, but I would've never come up with this way.

OK. I think I actually understood that. Is this the solution?
P(A) = .02, P(B) = .02, P(A`) = 1-.02=.98, P(B`)= .98

P(A∩B)= P(A)xP(B)= (.02)x(.02) = .0004
P(A`∩B`)= P(A`)xP(B`)= (.98)x(.98)= .9604
P(A∪B)= P(A)xP(B`)+P(A`)xP(B)= (.0196)+(.0196)= .0392

The probability that both alternators fail is .04%
The probability that neither alternator fails is 96%
The probability that one or the other fails is 3.9%

I tried using the binomial formula but got a little lost. Is the correct formula? (n/x)(p)xpower(1-p)n-xpower?

I did find it easier to calculate but my answers were off a little from what I have above and if the equation is n/x then would 0/2(neither of the alternators fail) be O?

If you use the binomial distribution, then you have:

X~B(0.02, 2)

P(X = 1) = 2C1(1- 0.02)^1(0.02)^1 = 0.0392

For part (b), 0C1(0.02)^0(1-0.02)^2 = 0.98^2.

0C1 is 1.