Question #5013

each time Mae and Richard play chess, Mae has probability of 4/5 that she wins. If they play 5 games, determin the probability that a) Mae wins 3 games b) Mae wins either 4 or 5 games.

Expert's answer

Each time Mae and Richard play chess, Mae has probability of 4/5 that she wins. If they play 5 games, determine the probability that

a) Mae wins 3 games

It means Mae wins three games and loose two games, so

P = (4/5)³(1-4/5)² = (4/5)³(1/5)² = 0.02048

b) Mae wins either 4 or 5 games

The probability Mae wins 4 games (win four games and loose one):

P1 = (1-4/5)(4/5)^4 = 0.08192.

The probability Mae wins 5 games:

P1 = (4/5)^5 = 0.32768.

So, the desired probability is

P = P1+P2 = 0.08192 + 0.32768 = 0.4096.

a) Mae wins 3 games

It means Mae wins three games and loose two games, so

P = (4/5)³(1-4/5)² = (4/5)³(1/5)² = 0.02048

b) Mae wins either 4 or 5 games

The probability Mae wins 4 games (win four games and loose one):

P1 = (1-4/5)(4/5)^4 = 0.08192.

The probability Mae wins 5 games:

P1 = (4/5)^5 = 0.32768.

So, the desired probability is

P = P1+P2 = 0.08192 + 0.32768 = 0.4096.

## Comments

## Leave a comment