# Probability of two or more independent events

I came up with the following answers for this problem by calculating the chances of two or more independent events by multiplying the chances. Is my approach correct?

Lebron, Duane and Stephen are in 3-point shoot-out match, round-robin style.

The probability that LeBron will beat Duane is 35% The probability that LeBron will beat Stephen is 60% The probability that Duane will beat Stephen is 55%

What is the probability Stephen will lose both rounds of competition? 33% What is the probability Duane will win against LeBron and lose against Stephen? 29% What is the probability Stephen will win exactly one round of competition? 18%

• How did you calculate the probability Stephen will win exactly one round of competition? Mar 3, 2020 at 7:54

## 2 Answers

What is the probability Stephen will lose both rounds of competition?

Probability of Stephen losing to Lebron $$\times$$ Probability of Stephen losing to Duane

$$0.6 \times 0.55 = 0.33$$

What is the probability Duane will win against LeBron and lose against Stephen?

Probability of Duane winning against Lebron $$\times$$ Probability of Daune losing to Stephen

$$(1 - 0.35) \times (1-0.55) = 0.2925$$

What is the probability Stephen will win exactly one round of competition?

Two cases:

1. Probability of Stephen winning against Lebron $$\times$$ Probability of Stephen losing to Daune

2. Probability of Stephen winning against Daune $$\times$$ Probability of Stephen losing to Lebron

Add these together.

$$((1 - 0.6) \times 0.55) + (0.6 \times (1-0.55)) = 0.49$$

• Thank you eyl327. Yes, this is the way I did the first two, except the last one. Mar 3, 2020 at 8:16

I don't agree with the last answer. Remember there is two possiblities of Steph to win exactly one round.

$$P(\text{stephen winning one round}) \\= P(\text{stephen beating lebron, and losing to duane}) + P(\text{stephen losing to lebron, and winning over duane})\\=0,55 \cdot (1-0,60)+(1-0,55)\cdot 0,6=0,49$$