Question 1192063: A boy named Curt wants to join the chess club. He has to play 3 games with players Andrew(A) and Ben(B) and NOT lose two games in a row. Chances of Curt winning against Andrew are 0.44 and against Ben - 0.43. Games cannot end in a draw and their outcome isn't influenced by the outcome of any previous games. Curt can choose with whom to play in two ways - A-B-A or B-A-B. What is the probability of joining the chess club if Curt chooses to play A-B-A? B-A-B?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
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:- P(not A,B,A) = P(not A)*P(B)*P(A) = 0.56*0.43*0.44 = 0.105952
- P(A,not B,A) = P(A)*P(not B)*P(A) = 0.44*0.57*0.44 = 0.110352
- P(A,B,not A) = 0.105952 through similar calculations as the first scenario.
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- probability winning all three games = 0.083248
- probability winning exactly 2 games = 0.322256
- probability winning exactly 1 game, and still able to join = 0.134848
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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