Question 1203673
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A softball team has three pitchers, A, B, and C, with winning {{{highlight(cross(percentages))}}} <U>rates</U> of 0.4, 0.6, and 0.8 respectively. 
These pitchers pitch with frequency 2,3 and 5 out of every 10 games, respectively 
In other words for a randomly selected game P(A) = 0.2 P(B) = 0.3, and P(C) = 0.5:
(a) Find P(team wins game) = P(W)
(b) P(A pitched game | team won) = P(A|W)
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<pre>
                     <U>Part (a) solution</U>


Pitcher A contributes to winning  {{{(2/10)*0.4}}} = 0.2*0.4 = 0.08.

Pitcher B contributes to winning  {{{(3/10)*(0.6)}}} = 0.3*0.6 = 0.18.

Pitcher C contributes to winning  {{{(5/10)*(0.8)}}} = 0.5*0.8 = 0.4.


The probability to win a game is the sum of these partial probabilities 

    P(W) = 0.08 + 0.18 + 0.4 = 0.66,

since their contributions are disjoint.



                     <U>Part (b) solution</U>



It is the conditional probability 


    P(A|W) = {{{P(A_and_W)/P(W)}}} = {{{0.08/0.66}}} = 0.1212  (rounded).    <U>ANSWER</U>
</pre>

Solved, with complete explanations.



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I replaced the word "percentage" in the post, since there is NO any percentage there.


Simply the problem's creator likes to use "smart" scientific terms, but makes it lubberly.