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3 light bulbs are picked without replacement from a bucket with 15 bulbs of which 5 are broken.

What is the probability that...

1.) None of the bulbs picked are broken.

2.) Exactly one of the bulbs picked is broken.

3.) At least one of the bulbs picked is broken.

For 1.) I did 5 *(5 0)/(15 3) = 1/91

2.) 5 * (5 1)/(15 3) = 5/91

3.) 1 - (10 3)/(15 3) = 67/91

Are these done correctly?

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  • $\begingroup$ How many are we picking? $\endgroup$ Mar 12, 2014 at 2:52
  • $\begingroup$ We're picking 3. $\endgroup$
    – rojas777
    Mar 12, 2014 at 2:53

4 Answers 4

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1.) The first one should look like this: $$\frac{\binom{5}{0}\binom{10}{3}}{\binom{15}{3}}$$ You want to pick no broken ones, so all 3 should come from the unbroken pile.

2.) You should modify my answer to 1.) to get this answer.

3.) The opposite of at least one broken bulb is no broken bulbs, right? So if an event $A$ has a probability of $P(A)$, then the probability of it not occurring is $1-P(A)$. You can use this idea and the answer to 1.) to get the desired answer.

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The total number of ways to pick 3 lightbulbs from a pile of 15 is (15C3), so that'll be your denominator for the following parts.

1) In the first case, you want to choose 3 lightbulbs out of the 10 functional ones, and 0 lightbulbs out of the 5 broken ones. You take their product because these events are not mutually exclusive in this context, and divide it by the total number of combinations of choosing 3 lightbulbs.s

Answer: ((10C3)*(5C0)) /(15C3)

2) You modify the solution to part 1 to get the following:

Answer: ((10C2)*(5C1)) /(15C3)

3) There are two ways to go about this question. You can either add the probability of choosing exactly one broken lightbulb OR exactly 2 broken lightbulbs OR exactly three broken lightbulbs. This method will result in the following formula:

Answer: ((10C2)(5C1) + (10C1)(5C2) + (10C0)*(5C3)) /(15C3)

Another method which would result in an equivalent solution is determining the probability of selecting no broken lightbulbs (calculated in part1), and subtract it from 1.0.

Answer: 1.0 - ((10C3)*(5C0)) /(15C3)

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1) (10/15)(9/14)(8/13)

2) (5/15)(10/14)(9/13)

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You can split the 15 bulbs into two boxes, the one with the 5 broken bulbs and the other one with the 10 non-broken bulbs.

Now if you are to find the probability to pick exactly n broken bulbs you first need to find the number of possibilities for that picking. This is just

(5 n)*(10 3-n)

as you can think of it as picking n bulbs of the box with defect bulbs and the rest of the 3 pickings need to be non-broken bulbs.

So the probability is just

p(n)= (5 n)*(10 3-n) /(15 5)

Now you just need to put n (the amount of broken bulbs into that formula)

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