Please post the answer tot he question. I have some ideas but i'm getting multiple answers.

Is n = 4? Pl. post OA.

n=3?

probability of getting one defective and one non defective bulb = (prob of getting first one defective and second one non-defective) + (probability of getting the first one non-defective and the second one defective)
--> (n/10 * 10-n/9) + (10-n /10 * n/9) = 7/15
--> factorize to solve for n to give n = 7 or 3, since n <5 hence n =3

should the question be asking which n gives you 7/15?

10C2 = 45 given the total number of ways to pick 2 bulbs out of 10

6G4D = 24 pairs of 1G and 1D -> 24/45 = 8/15
7G3D = 21 pairs of 1G and 1D -> 21/45 = 7/15
8G2D = 16 pairs of 1G and 1D -> 16/45
9G1D = 9 pairs of 1G and 1D -> 9/45

confused

Here's the question cut and pasted:

There are 10 light bulbs and fewer than half are defective. If n is the number of defective light bulbs, what is the value of n if the probability that one of the bulbs will be defective and the other will not be = 7/15?

Ok, with that said, this is a difficult
simultaneous probability question.

The question is giving you two scenarios. The first one, you pick a defective light bulb and THEN a non-defective light bulb. Consider this P(A) * P(B). In the second scenario, you pick a good light bulb on the first try and THEN a defective light bulb. Consider this P(B) * P(A).

Now that you have the two equations, you will need to ADD them together. Why? Because either/OR will work, and whenever you see the word "or", think: addition. (When you see the word "and", think multiply)

Let's set up the full equation:

[P(A) * P(B)] + [P(B) * P(A)]


Let's try n=3.

(3/10 * 7/9) + (7/10 * 3/9)
= 21/90 + 21/90
= 42/90
= 7/15

Why is the denominator a "9" in the second piece of each multiplication? Well, if you take 1 light bulb out of a batch of 10, then on the second try, how many light bulbs do you have to choose from? Right, you have 9. Consider this a probability without replacement simultaneous equation.

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