Question

wednesday, november 12th, 2025
bellwork
probability review
a team plays two games. their
chance of winning the 1st game is
60%, chance of winning the 2nd is
90%. determine:

1. p(win both games)
2. p(lose both games)
3. p(win exactly one game)
4. p(win at least one game)

Explanation:

Step1: Calculate probabilities of winning and losing each game

Let \(P(W_1) = 0.6\) (probability of winning 1st game), \(P(L_1)=1 - 0.6=0.4\) (probability of losing 1st game)
\(P(W_2) = 0.9\) (probability of winning 2nd game), \(P(L_2)=1 - 0.9 = 0.1\) (probability of losing 2nd game)

Step2: Find \(P(\text{win both games})\)

Since the games are independent events, \(P(W_1\cap W_2)=P(W_1)\times P(W_2)\)
\(P(W_1\cap W_2)=0.6\times0.9 = 0.54\)

Step3: Find \(P(\text{lose both games})\)

\(P(L_1\cap L_2)=P(L_1)\times P(L_2)\)
\(P(L_1\cap L_2)=0.4\times0.1=0.04\)

Step4: Find \(P(\text{win exactly one game})\)

\(P(\text{win exactly one game})=P(W_1\cap L_2)+P(L_1\cap W_2)\)
\(P(W_1\cap L_2)=0.6\times0.1 = 0.06\)
\(P(L_1\cap W_2)=0.4\times0.9=0.36\)
\(P(\text{win exactly one game})=0.06 + 0.36=0.42\)

Step5: Find \(P(\text{win at least one game})\)

Using the formula \(P(\text{win at least one game})=1 - P(\text{lose both games})\)
\(P(\text{win at least one game})=1-0.04 = 0.96\)

Answer:

1. \(0.54\)
2. \(0.04\)
3. \(0.42\)
4. \(0.96\)