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?
Probability problem on light bulbs
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crackthetest
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life is a test
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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
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
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jzdchou
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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
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
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David_IBaspirant
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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.
If you like this post, I am only a student. I took the Veritas GMAT prep course and gotta vouch for them. They review combinatorics and probabilities in depth. Email me for info. I'm testing this Friday for the GMAT, and ready to roll. Best![/b]
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.
If you like this post, I am only a student. I took the Veritas GMAT prep course and gotta vouch for them. They review combinatorics and probabilities in depth. Email me for info. I'm testing this Friday for the GMAT, and ready to roll. Best![/b]
