SOLUTION: A manufacturer of tire found out that 2 % of his production was defective. If you have just bought a new car with 4 of those tires , what is the probability that one of these tire

Algebra ->  Probability-and-statistics -> SOLUTION: A manufacturer of tire found out that 2 % of his production was defective. If you have just bought a new car with 4 of those tires , what is the probability that one of these tire      Log On


   



Question 263823: A manufacturer of tire found out that 2 % of his production was defective.
If you have just bought a new car with 4 of those tires , what is the probability that one of these tires is defective.
thanks for your help

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
2% of production is defective.

the probability of any tire being defective is 2%.

this means that the probability of not getting a defective tire is 98%

the probability of getting 0 defective tires out of 4 tires = .98^4 = .92236815.

the probability of getting at least 1 defective tire is therefore 1 - .92236815 = .07763184.

here's how it works.

the probability of getting exactly 0 defective tires is 1 * .98^4 = :
.92236815.
the probability of getting exactly 1 defective tire is 4 * .02 * .98^3 = :
.07529536
the probability of getting exactly 2 defective tires is 6 * .02^2 * .98^2 = :
.00230496
the probability of getting exactly 3 defective tires is 4 * .02^3 * .98 = :
.00003136
the probability of getting exactly 4 defective tires is 1 * .02^4 = :
.00000016

sum up all those probabilities and you get 1.

this is what you should get.

I summed up the numbers stored in the calculator that were not rounded so I got exactly 1. If you add up the displayed numbers, you will get close to 1 but not right on.

your answer is:

the probability of getting at least 1 defective tire is .07763184.

the probability of getting exactly 1 defective tire is .07529536

the implicit assumption in all of this is that the manufacturer is making a very large number of tires, so that the probability can be calculated assuming replacement rather than not assuming replacement.

the multipliers in each operation are based on combination theory.

the possible number of ways of getting a set of 0 out of 4 is 1
the possible number of ways of getting a set of 1 out of 4 is 4
the possible number of ways of getting a set of 2 out of 4 is 6
the possible number of ways of getting a set of 3 out of 4 is 4
the possible number of ways of getting a set of 4 out of 4 is 1

assume your set is abcd.

there is only 1 way of getting a set of 0 out of this.
you just don't draw any.

there is 4 ways of getting a set of 1 out of this.
you can get a, b, c, or d.

there is 6 ways of geting a set of 2 out of this.
you can get ab, ac, ad, bc, bd, cd

there is 4 ways of getting a set of 3 out of this.
you can get abc, abd, acd, bcd

there is 1 way of getting a set of 4 out of this.
you can get abcd only.

these are combinations, not permutations. order is not important. abc is considered the same set as acb, bac, bca, cab, cba.