Binomial Distribution: Finding Probability with Trials, Success, and X Value

[TyErd]

Homework Statement

I've uploaded a picture of the question.
I need help in identifying the correct number of trials, probability of success and the X value(number of successes)

Homework Equations

i'm using the binomial distribution function on the calculator but I've attached the formula just in case

The Attempt at a Solution

I used n=7, p=7/15 and X=7 which yields me a probability of 0.00482 which is incorrect. The correct answer is 0.000155 but not sure how.

 

[LCKurtz]

Homework Statement

I've uploaded a picture of the question.
I need help in identifying the correct number of trials, probability of success and the X value(number of successes)

Homework Equations

i'm using the binomial distribution function on the calculator but I've attached the formula just in case

The Attempt at a Solution

I used n=7, p=7/15 and X=7 which yields me a probability of 0.00482 which is incorrect. The correct answer is 0.000155 but not sure how.

The binomial distribution is inappropriate for this problem. You aren't doing independent trials with replacement. Think about how many ways you can select 7 cars from the 15 and how many ways you can select the 7 fwd cars.

 

[Dick]

That's not a correct formula to use. How many ways are there to choose 7 cars from 15? Only one of those choices gives you all four wheel drives.

 

[TyErd]

ahh, that's why i keep getting it wrong. Thanks, but if its not binomial, what is it?

 

[Dick]

It's combinatorics. How many ways to select 7 objects from 15 objects?

 

[Ray Vickson]

ahh, that's why i keep getting it wrong. Thanks, but if its not binomial, what is it?

It is the so-called hypergeometric distribution.

Note: instead of a combinatorial argument there is another way to get the correct answer. The probability that the first car is fwd is 7/15; that leaves 14 cars, of which 6 are fwd. So (given the first is fwd) the probability that the second is fwd is 6/14, etc, etc.

RGV