Question

Suppose George loses 42​% of all chess games. ​

(a) What is the probability that George loses two chess games in a​ row? ​

(b) What is the probability that George loses four chess games in a​ row?

​(c) When events are​ independent, their complements are independent as well. Use this result to determine the probability that George loses four chess games in a​ row, but does not lose five in a row. ​

(a) The probability that George loses two chess games in a row is ? ​(Round to four decimal places as​ needed.)

​(b) The probability that George loses four chess games in a row is ?. ​(Round to four decimal places as​ needed.) ​

(c) The probability that George loses four chess games in a​ row, but does not lose five in a row is ?. ​(Round to four decimal places as​ needed.)

Answer 1

P(loss) = 42% = 0.42

P(win) = 58% = 0.58

Since each game played is independent of the next one, probability of  losing is the same in each play.

Answer a.

P(losing two games in a row) = P(loss 1 and loss 2) = P(loss 1)*P(loss 2)

P(losing two games in a row) =

Answer b.

P(losing four games in a row) = P(loss 1 and loss 2 and loss 3 and loss 4)

P(losing four games in a row) = P(loss 1)*P(loss 2)*P(loss 3)*P(loss 4)

P(losing four games in a row) =

Answer c.

P(losing four games in a row but does not lose five in a row) =

P(win 1)*P(loss 2)*P(loss 3)*P(loss 4)*P(loss 5) +  P(loss 1)*P(loss 2)*P(loss 3)*P(loss 4)*P(win 5)

P(losing four games in a row but does not lose five in a row) =