Answer:
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
