Question 1192063
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Part 1)  A-B-A


P(A) = probability Curt wins against Andrew
P(A) = 0.44


P(B) = probability Curt wins against Ben
P(B) = 0.43


P(A,B,A) = probability of winning against Andrew, Ben, then Andrew again in that order.
P(A,B,A) = P(A)*P(B)*P(A) which is valid because events are independent
P(A,B,A) = 0.44*0.43*0.44
P(A,B,A) = 0.083248
This is the scenario where Curt wins all three games in a row.


Let's say he's not so lucky and loses exactly one game.
We have three possibilities:<ul><li>P(not A,B,A) = P(not A)*P(B)*P(A) = 0.56*0.43*0.44 = 0.105952</li><li>P(A,not B,A) = P(A)*P(not B)*P(A) = 0.44*0.57*0.44 = 0.110352</li><li>P(A,B,not A) = 0.105952 through similar calculations as the first scenario.</li></ul>Add those results:
0.105952+0.110352+0.105952 = 0.322256


The probability of Curt losing exactly one game is 0.322256, which is the same as the probability of Curt winning exactly 2 games.


Now let's further say Curt's luck gets worse.
We'll consider the possibility of him losing 2 games.
The only way this can happen, and he still joins the club, is that he loses the first and last games. Any other scenario means he can't join the club.
P(not A, B, not A) = P(not A)*P(B)*P(not A)
P(not A, B, not A) = 0.56*0.43*0.56
P(not A, B, not A) = 0.134848


Let's recap<ul><li>probability winning all three games = 0.083248</li><li>probability winning exactly 2 games = 0.322256</li><li>probability winning exactly 1 game, and still able to join = 0.134848</li></ul>The final step is to sum those three scenarios:
0.083248 + 0.322256 + 0.134848


If Curt goes with the A-B-A format, then the chances of him joining the chess club is 0.540352


So it's a 54.0352% chance.


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The steps for the B-A-B case will follow the same idea as the last section. 
I'll let the student handle this to get practice.
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