If a truck has 18 tires, and the probability of each tire blowing out is 0.03?

Question

What is the probability that exactly one tire blows out? Two?

I've tried 0.97^17 times 0.03 for one tire, and got 1.787%. The correct answer is 32%.

Answer 1

This is a binomial distribution problem.

(A) The probability of exactly one blowout is the product of:

(i) the probability of one blowout (0.03^1)
(ii) the probability of 17 non-blowouts (0.97^17)
(iii) the number of combinations for one blowout (18!/(1! * (18-1)!) = 18)

0.03^1 * 0.97^17 * (18!/(1! * (18-1)!) =
0.03 * 0.596 * 18 =
0.322

(B) The probability of exactly two blowouts is the product of:

(i) the probability of two blowouts (0.03^2)
(ii) the probability of 16 non-blowouts (0.97^16)
(iii) the number of combinations for two blowouts (18!/(2! * (18-2)!) = 18*17/2)

0.03^2 * 0.97^16 * (18!/(2! * (18-2)!) =
0.0009 * 0.614 * 153 =
0.0846

So, it's about 32% for one blowout, and 8% for two.

Answer 2

There's a little conceptual trick here: in order to calculate those probabilities, you need to consider what happens to all the tires, not just the one that blows out. (The second trick here is that, in order to get an answer from your information, you have to assume that a blowout on one tire doesn't affect any of the others. Since this is obviously false, given that one blowout will put more stress on the other tires, you should take your calculated answers with several tons of salt.)

To explain what I mean, let's take a case you didn't mention: what's the chance of exactly five tires blowing out?

The first thing to note here is that nobody said *which* five tires blow out. This means we'll have a multiplying factor coming in from the number of possible combinations of five tires. In fact, that multiplying factor is *called* a combination - more specifically, a combination without repetition, since you won't have the same tire blow twice, but if you say 'combination', people will know what you mean. At least, if they're statisticians - you may have to say "binomial coefficient" to other people. Anyway, in our example, you need "18 choose 5" - that is, the number of different sets of 5, not worrying about order, that you can get from 18. The formula should be in your textbook - it's certainly on the webpage linked in the first source.

The second number you need is the chance of five *specific* tires blowing out of the eighteen - and only those five. (Remember, you want 'exactly five', not 'at least five'.) Since the chance of each individual blowout is 0.03, one term will be (0.03)^5 - the odds of five independent blowouts. The other term will be the 'exactly five' part - that is, (1-0.03)^(18-5). This is the chance of the other thirteen tires *not* blowing out.

In any case, multiply all these numbers - 18 choose 5, (0.03)^5, and (1-0.03)^(18-5) - and that should give you the answer to the example: approximately 0.0001401. You can probably expect bigger numbers for one tire and two tires. (And you can probably guess which numbers should be replaced with 1 and 2 in the formula.)