Help Binomial Distribution: Statistics for M.E's

[geno678]

Help! Binomial Distribution: Statistics for M.E's

Homework Statement

Four wheel bearings are to be replaced on a company vehicle. The mechanic has selected the four replacement parts from a large supply bin in which 10% of the bearings are defective and will fail within the first 100 miles. What is the probability that the company vehicle will have a breakdown due to defective wheel bearings?

Homework Equations

Binomial Distribution Formula.

Pn(x) = C(n,x) (P^x) (q^(n-x)) = n! / ( (x!) * (n -x)! ) * (P^x) * (q^n-x)

I tried Latex but for some reason it wouldn't work properly.

The Attempt at a Solution

I know that q probability of failure =.1 q = p - 1
and probability of succes =.9


n # number of trials = 4
X # of successes = ?

I can plug in the values that's not the problem for me.

The main problem is the number of trials and successes. I don't know if I'm right.

A mechanic chooses 4 new bearings out of a large bin. So I'm assuming the number of trials is 4. But I don't know what the # of successes are. I have a 90% chance that the bearings will be fine, so what does this mean? Unless I have more than 4 trials.

 

[geno678]

Ok I think I got it, they said 10 percent were defective out of a large bin. They chose 4.
If I multiply 4 * .1 = .4 that is 4 out of 10 that are defective. Then this must mean 5/10 are successful. That means my p isn't .9 it is .5

 

[LCKurtz]

No. The probability that a selected bearing is good is .9. Since the supply of bearings is large you can assume that whether or not a bearing chosen is good or bad doesn't affect the probability for the next bearing selected. So figure the choices are independent.

So what is the probability all four are good?

If they aren't all good, there will be an accident. What is the probability of that given the answer to my first question?

 

[geno678]

The probability that all four are good is. (1/4)*(1/4)*(1/4)*(1/4) = 1/256 = .4 chance that all four are good

 

[LCKurtz]

The probability that all four are good is. (1/4)*(1/4)*(1/4)*(1/4) = 1/256 = .4 chance that all four are good

Where are you getting the 1/4 from? The problem says that 90% of them are good.

 

[geno678]

Ok wait a minute. So 90% probability for each bearing that it is good.

Then I guess for four bearings it would be (9/10)*(9/10)*(9/10)*(9/10) = 65.61%

 

[geno678]

For some reason I was thinking you had 1 out 4 chances of a bearing being right, that's why I said 1/4. My mistake.

 

[geno678]

If they aren't all good it will be. (.1)*(.1)*(.1)*(.1) = .01%

 

[geno678]

Am I way off??

 

[geno678]

I get it. It's just 90%. Because it's consistent for every trial.

 

[geno678]

Ok I get it now. Their are 8 trials because, 1 wheel bearing can be good or bad. That's 2 trials for 1 wheel bearing. For four wheel bearings, you get 8 trials. And the number of successes are 4.

 

[LCKurtz]

Ok wait a minute. So 90% probability for each bearing that it is good.

Then I guess for four bearings it would be (9/10)*(9/10)*(9/10)*(9/10) = 65.61%

Yes. That is the probability that all four are good. Now how can you figure out the probability that they aren't all good from that? That is what you are looking for.

 

[geno678]

Ok I plugged in all of my variables, but the equation is set up to find the success of the parts. Do I need to switch up some variables?