Question 1203673
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Part (a)


We have this information provided
P(A) = 0.2
P(B) = 0.3
P(C) = 0.5
and
P(W given A) = 0.4
P(W given B) = 0.6
P(W given C) = 0.8


Something like "P(W given A)" means "the probability of a win when we know 100% that pitcher A played".
P(W given A) = 0.4 can be rewritten to "P(W) = 0.4 when pitcher A plays"


This conditional probability formula
P(A given B) = P(A and B)/P(B)
rearranges to
P(A and B) = P(A given B)*P(B)


Then use the law of total probability
P(W) = P(W and A) + P(W and B) + P(W and C) 
P(W) = P(W given A)*P(A) + P(W given B)*P(B) + P(W given C)*P(C)
P(W) = 0.4*0.2 + 0.6*0.3 + 0.8*0.5
P(W) = <font color=red size=4>0.66</font>



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Part (b)


I'll use Bayes Theorem
P(A given B) = P(B given A)*P(A)/P(B)


So,
P(A given W) = P(W given A)*P(A)/P(W)
P(A given W) = 0.4*0.2/0.66
P(A given W) = 0.08/0.66
P(A given W) = 8/66
P(A given W) = <font color=red>4/33</font>
P(A given W) = <font color=red size=4>0.121212...</font> where the "12" repeats forever
Round this however needed. 


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Another approach for Part (a)


Let's say the team plays 100 games.


Pitcher A does 20% of those games, so they pitch for 0.2*100 = 20 games.
Pitcher B then does 0.3*100 = 30 games
Pitcher C then does 0.5*100 = 50 games


Here's the table so far
<table border = "1" cellpadding = "5"><tr><td></td><td>Win</td><td>Loss</td><td>Total</td></tr><tr><td>A pitches</td><td></td><td></td><td>20</td></tr><tr><td>B pitches</td><td></td><td></td><td>30</td></tr><tr><td>C pitches</td><td></td><td></td><td>50</td></tr><tr><td>Total</td><td></td><td></td><td>100</td></tr></table>
For the "A pitches" row, we'll have 8 wins and 12 losses.
This is because pitcher A has a 40% success rate, so they win 0.4*20 = 8 games and lose the other 20-8 = 12 games.
The other rows are filled out in a similar fashion.


We'll get this completed table
<table border = "1" cellpadding = "5"><tr><td></td><td>Win</td><td>Loss</td><td>Total</td></tr><tr><td>A pitches</td><td>8</td><td>12</td><td>20</td></tr><tr><td>B pitches</td><td>18</td><td>12</td><td>30</td></tr><tr><td>C pitches</td><td>40</td><td>10</td><td>50</td></tr><tr><td>Total</td><td>66</td><td>34</td><td>100</td></tr></table>
There are 66 wins out of 100 games total. 
Therefore, P(W) = 66/100 = <font color=red>0.66</font>


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Another approach for Part (b)


Refer to the table in the previous section.


There are 66 wins. Of those wins, 8 are when pitcher A was in the game.
P(A given W) = P(A and W)/P(W)
P(A given W) = 8/66
P(A given W) = <font color=red>4/33</font>
P(A given W) = <font color=red>0.121212...</font>
We focus on the "win" column because we know 100% that a win occurred. We ignore the loss column.
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