Find the probability distribution

[Sucks@Physics]

Homework Statement

A box of 10 flashbublbs contains 3 defective bublbs. A random sample of 2 is selcted and tested. Let X be the radom variable associated with the number of defective bulbs in the sample.

A) Find the probability distribution of X.

B) Find the expected number of defective bulbs in a sample.

I did this problem, I got 7/15 probability of there being no defective light bulbs, 1/15 of there being 2, but I got 7/30 of there being 1. The 7/30 is wrong but I don't know where I'm going wrong, please give me some help. Thanks in advance.

Homework Equations

The Attempt at a Solution

7/10*6/9 = 7/15
7/10*3/9 = 7/30
3/10*7/9 = 1/15

 

[tiny-tim]

7/10*6/9 = 7/15
7/10*3/9 = 7/30
3/10*7/9 = 1/15

Hi Sucks@Physics !

(in the last line, you wrote a 7 instead of a 2 )

As you've probably noticed, the three probabilities should add up to 1.

Your 7/30 is the probability of the first one being defective, and the second one good … but you also need the probability of the first one being good, and the second one defective … that doubles it!

 

[Architeuthis Dux]

Firstly, it is important to note that the probability distribution of X in this scenario follows a binomial distribution, since there are only two possible outcomes (defective or non-defective) for each tested bulb and the sample size is fixed at 2.

A) To find the probability distribution of X, we can use the binomial probability formula: P(X=k) = nCk * p^k * (1-p)^(n-k), where n is the sample size, p is the probability of success (in this case, finding a defective bulb), and k is the number of successes.

In this problem, n=2, p=3/10 (since there are 3 defective bulbs out of 10 total bulbs), and k can take on values of 0, 1, or 2. Therefore, the probability distribution is:

P(X=0) = 2C0 * (3/10)^0 * (7/10)^2 = 49/100

P(X=1) = 2C1 * (3/10)^1 * (7/10)^1 = 42/100

P(X=2) = 2C2 * (3/10)^2 * (7/10)^0 = 9/100

B) To find the expected number of defective bulbs in a sample, we can use the formula: E(X) = n * p, where n is the sample size and p is the probability of success.

In this problem, n=2 and p=3/10. Therefore, the expected number of defective bulbs in a sample is 2 * 3/10 = 0.6. This means that, on average, we would expect to find 0.6 defective bulbs in a sample of 2 bulbs.

In your attempt at the solution, it seems that you have mixed up the probabilities for finding 0 or 1 defective bulbs. The correct probabilities for P(X=0) and P(X=1) are 49/100 and 42/100, respectively. The probability for finding 2 defective bulbs is correct at 9/100.