Question 1150084
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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
<table border = "1" cellpadding = "5">
<tr><td>X</td><td>P(X)</td></tr>
<tr><td>0</td><td>38/87</td></tr>
<tr><td>1</td><td>40/87</td></tr>
<tr><td>2</td><td>3/29</td></tr>
</table>



The fractions can be replaced with their decimal approximations
<table border = "1" cellpadding = "5">
<tr><td>X</td><td>P(X)</td></tr>
<tr><td>0</td><td>0.43678</td></tr>
<tr><td>1</td><td>0.45977</td></tr>
<tr><td>2</td><td>0.10345</td></tr>
</table>
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
<table border = "1" cellpadding = "5">
<tr><td>X</td><td>P(X)</td></tr>
<tr><td>0</td><td>38/87</td></tr>
<tr><td>1</td><td>40/87</td></tr>
<tr><td>2</td><td>3/29</td></tr>
</table>


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
<table border = "1" cellpadding = "5">
<tr><td>X</td><td>P(X)</td><td>X*P(X)</td></tr>
<tr><td>0</td><td>38/87</td><td>0*(38/87) = 0</td></tr>
<tr><td>1</td><td>40/87</td><td>1*(40/87) = 40/87</td></tr>
<tr><td>2</td><td>3/29</td><td>2*(3/29) = 6/29</td></tr>
</table>


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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