Alex and bob are playing 5 chess games. Alex is 3 times more likely to win than bob. What is the probability that both of them will win at least 2 games? I'll give a brainliest to whoever answers it correctly first.

12.2%

Step-by-step explanation:

Lets firstly find out the probabilities for Alex and Bob. We know that the prb that Alex wins is 3 times the prob for Bob, so:

P(A) = 3P(B)

And, as these values are probabilities they must sum 1:

P(A) + P(B) = 1

Replacing the first equation in the second:

3P(B) + P(B) = 1

4P(B) = 1

Dividing both sided by 4:

4P(B)/4 = 1/4

P(B) =1/4

So, the probability for Bob is 1/4. Then the probability for Alex is 3*1/4 = 3/4

Now we have to find the probability that both win at least two games. Lets think about the possible combination of cases. As they play 5 games and both have to win at least to there is a remaining game that cane variate, this is, can be won by Alex or Bob. So, we can have:

AABBB

AABBA

Where A is "Alex win" and B is "Bob win" (here we do not pay attention to the order).

Thus, we have to find the probability that Alex wins 3 games and Bob 2 games, and the probability that Alex wins 2 games and Bob 3 games. As the games are independent, i.e., the result of one games does not affect the followings we can just multiply the probabilities:

P(3A and 2B) = (3/4)*(3/4)*(3/4)*(1/4)*(1/4) = 0.42*0.06 = 0.026

P(2A and 3B) = (3/4)*(3/4)*(1/4)*(1/4)*(1/4) = 0.56*0.016 = 0.096

So,

P (at least 2) = P(3A and 2B) + P(2A and 3B) = 0.026 + 0.096 = 0.122 = 12.2%

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