Find the probability distribution

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SUMMARY

The discussion focuses on calculating the probability distribution for a random variable X, representing the number of defective bulbs in a sample of 2 from a box containing 10 bulbs, of which 3 are defective. The correct probabilities are determined to be 7/15 for zero defective bulbs, 1/5 (or 2/15) for two defective bulbs, and 7/15 for one defective bulb after correcting the initial miscalculation. The key takeaway is that the probabilities must sum to 1, and both arrangements of defective and non-defective bulbs must be considered in the calculations.

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Sucks@Physics
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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
 
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Sucks@Physics said:
7/10*6/9 = 7/15
7/10*3/9 = 7/30
3/10*7/9 = 1/15

Hi Sucks@Physics ! :smile:

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

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! :smile:
 

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