Question 1150084: A box of 30 flashbulbs contains 10 defective bulbs. A random sample of 2 is selected and tested. Let X be the random 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 the sample.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Part A
X = number of defective bulbs
X is an integer chosen from the set {0, 1, 2}.
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If X = 0, then we have no defective bulbs, which means both bulbs are working.
The probability of choosing 2 working bulbs is (20/30)*(19/29) = 38/87
20/30 represents the probability of choosing one working bulb (20 working out of 30 total)
19/29 represents the probability of choosing a second working bulb (each value goes down by 1 becuse we dont replace whatever bulb was selected)
So if X = 0, then P(X) = 38/87
We'll have this in the table below
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When X = 1, we have one working bulb and one defective bulb
A = P(first bulb works) = 20/30 = 2/3
B = P(second bulb doesnt work) = 10/29
C = A*B = (2/3)*(10/29) = 20/87
D = P(first bulb doesnt work) = 10/30 = 1/3
E = P(second bulb works) = 20/29
F = D*E = (1/3)*(20/29) = 20/87
We get the same result each time so the order doesnt matter (if we get defective first or not)
C+F = (20/87)+(20/87) = 40/87 is the probability of having one bulb working and the other defective
If X = 1, then P(X) = 40/87
We'll have this in the table below
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Finally, if X = 2 then we have 2 defective bulbs
P(both defective) = P(first defective)*P(second defective) = (10/30)*(9/29) = 3/29
So if X = 2, then P(X) = 3/29
We'll have this in the table below
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The probability distribution looks like this
| X | P(X) |
| 0 | 38/87 |
| 1 | 40/87 |
| 2 | 3/29 |
The fractions can be replaced with their decimal approximations
| X | P(X) |
| 0 | 0.43678 |
| 1 | 0.45977 |
| 2 | 0.10345 |
38/87 = 0.43678
40/87 = 0.45977
3/29 = 0.10345
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Part B
Refer to the table created in part A
| X | P(X) |
| 0 | 38/87 |
| 1 | 40/87 |
| 2 | 3/29 |
Let's add on a new column called X*P(X). This column is the result of multiplying each X and P(X) value together for any given row
| X | P(X) | X*P(X) |
| 0 | 38/87 | 0*(38/87) = 0 |
| 1 | 40/87 | 1*(40/87) = 40/87 |
| 2 | 3/29 | 2*(3/29) = 6/29 |
Add up the results in that new column
0+40/87+6/29 = 40/87+18/87 = 58/87 = (29*2)/(29*3) = 2/3
Expected number of defective bulbs in the sample = 2/3 = 0.667
Keep in mind that having a fractional amount of defective bulbs does not make sense for a single sample; however, if you repeatedly pulled out 2 bulbs at random (do this say 1000 times), then you should expect on average to have 0.667 defective bulbs.
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