A truck maker finds out that 20% of its truck transmissions on a certain model are defective and tells dealers to have owners bring the trucks in. A local dealer sold 7 of the trucks and has two new transmissions on hand. What is the probability that they need to get more transmissions? I thought that 20% of 7 is 1.4 and they alreadt have 2 but someone else said it was related to binomials and I was confused. HELP!

From how I interpret it, I may be wrong, but what it's asking is" suppose the probability of a tranmission being defective is 20%. What is the probability that given 7 transissions, 2 are defective". I hope I have interpreted correctly.


Yes, the binomial:



Where p=0.20, q=1-p, n=7 and k=2


EDIT: see last post. After Nosnosna's interpretation, I believe that's correct.

You're looking for the probability that, of those seven, more than two will need to be replaced.

If 20% of all transmissions are faulty, that does not mean that 20% of transmissions from this small set are... it's entirely possible that all seven of these are, and it's just as possible that none of the seven are. Each transmission has a 20% chance of being faulty, and nothing about the others in the set changes that.

There are 8 possible outcomes for the number of transmissions that will need to be replaced: 0, 1, 2, 3, 4, 5, 6, and 7. Without a generalized form for this, you would simply calculate the probability of each of those which would require more ordering (that is, 3, 4, 5, 6, and 7) and add them together to get the overall probability of having to order more. Generalized forms of course make this a single step problem, of course :)

The formula galactus posted looks correct, but it's been too long since I've done anything much with probability to say for sure :)

Edit: On further reflection, I believe that formula is for the probability of 'k' failures. So to get the probability of more than 2 failures, you'd do a summation from k=3 to 7 over that formula.

Yes, I think you are correct nosnosna. After reading the problem, it seems since they have two on hand, they need to know the probability of needing more.

Then: