A chess club has 5 players with unequal skill. Laurie is the best player. The probability that she will win depends on who her opponent is, with probabilities of winning of .6, .7, .8 and .9, respectively, when playing against members Carrie, David, Paul and Rose. Whether she wins or loses is independent from one game to the next. a. What is the probability that Laurie will lose when she plays against Rose? b. If Laurie plays one game against each player, what is the probability that she does not lose any games? c. If Laurie plays one game against each player, what is the probability that she loses at least once?

a) P(R') = 0.1

b) Probability that she doesn't lose any game if she plays them all = 0.3024

c) If Laurie plays one game against each player, the probability that she loses at least once = 0.6976

Step-by-step explanation:

Let the probability that Laurie will win against Carrie, David, Paul and Rose be P(C), P(D), P(P) and P(R) respectively.

P(C) = 0.6

P(D) = 0.7

P(P) = 0.8

P(R) = 0.9

a) Probability that Laurie will lose when she plays against Rose = P(R') = 1 - P(R) = 1 - 0.9 = 0.1

b) If Laurie plays one game against each player, the probability that she does not lose any games = the probability that she wins all the games = P(C) × P(D) × P(P) × P(R) = 0.6 × 0.7 × 0.8 × 0.9 = 0.3024

c) If Laurie plays one game against each player, the probability that she loses at least once is a sum of a number of probabilities

Take note,

Probability that she wins, against each opponent

P(C) = 0.6

P(D) = 0.7

P(P) = 0.8

P(R) = 0.9

The probability that she loses against each opponent

P(C') = 1 - P(C) = 1 - 0.6 = 0.4

P(D') = 0.3

P(P') = 0.2

P(R') = 0.1

1) Probability that Laurie loses only once while playing the 4 of them

= [P(C') × P(D) × P(P) × P(R)] + [P(C) × P(D') × P(P) × P(R)] + [P(C) × P(D) × P(P') × P(R)] + [P(C) × P(D) × P(P) × P(R')]

= (0.4×0.7×0.8×0.9) + (0.6×0.3×0.8×0.9) + (0.6×0.7×0.2×0.9) + (0.6×0.7×0.8×0.1) = 0.4404

2) Probability that she loses twice playing against each of them

= [P(C') × P(D') × P(P) × P(R)] + [P(C') × P(D) × P(P') × P(R)] + [P(C') × P(D) × P(P) × P(R')] + [P(C) × P(D') × P(P') × P(R)] + [P(C) × P(D') × P(P) × P(R')] + [P(C) × P(D) × P(P') × P(R')]

= (0.4×0.3×0.8×0.9) + (0.4×0.7×0.2×0.9) + (0.4×0.7×0.8×0.1) + (0.6×0.3×0.2×0.9) + (0.6×0.3×0.8×0.1) + (0.6×0.7×0.2×0.1) = 0.2144

3) Probability that Laurie loses three times while playing against them

= [P(C') × P(D') × P(P') × P(R)] + [P(C') × P(D') × P(P) × P(R')] + [P(C') × P(D) × P(P') × P(R')] + [P(C) × P(D') × P(P') × P(R')]

= (0.4×0.3×0.2×0.9) + (0.4×0.3×0.8×0.1) + (0.4×0.7×0.2×0.1) + (0.6×0.3×0.2×0.1) = 0.0404

4) Probability that she loses to the 4 opponents in a game against each one

= [P(C') × P(D') × P(P') × P(R')] = 0.4×0.3×0.2×0.1 = 0.0024

If Laurie plays one game against each player, the probability that she loses at least once = 0.4404 + 0.2144 + 0.0404 + 0.0024 = 0.6976

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step-by-step explanation: