Question

1. a baseball team plays a best 2 of 3 game series. their chance of winning either of the first two games is 45%, but their chance of winning a 3rd game (if it occurs) is 80%. let x = the number of games won.
2. a soccer team plays two games. in each game, they have a 60% chance of winning, 30% chance of losing, and 10% chance of tying. let x = the number of games won.

Explanation:

1. For the baseball - team problem (Question 5):

• Find the probability distribution of \(X\) (the number of games won):
• Case 1: \(X = 0\)
• # Explanation:

Step1: Lose first two games

The probability of losing the first game is \(1 - 0.45=0.55\), and the probability of losing the second game is also \(0.55\). Since these are independent events, the probability \(P(X = 0)=(1 - 0.45)\times(1 - 0.45)=0.55\times0.55 = 0.3025\).

• Case 2: \(X = 1\)
• # Explanation:

Step1: Win - lose scenario

The probability of winning the first game and losing the second is \(0.45\times(1 - 0.45)=0.45\times0.55\). The probability of losing the first game and winning the second is \((1 - 0.45)\times0.45\). Since these are mutually - exclusive events, \(P(X = 1)=0.45\times(1 - 0.45)+(1 - 0.45)\times0.45=2\times0.45\times0.55 = 0.495\).

• Case 3: \(X = 2\)
• # Explanation:

Step1: Win first two games

The probability of winning the first two games is \(0.45\times0.45 = 0.2025\).

Step2: Win first, lose second, then win third

The probability of winning the first game, losing the second, and then winning the third is \(0.45\times(1 - 0.45)\times0.8=0.45\times0.55\times0.8 = 0.198\).

Step3: Lose first, win second, then win third

The probability of losing the first game, winning the second, and then winning the third is \((1 - 0.45)\times0.45\times0.8=0.55\times0.45\times0.8 = 0.198\).
\(P(X = 2)=0.2025+0.198 + 0.198=0.6\).

1. For the soccer - team problem (Question 6):

• Find the probability distribution of \(X\) (the number of games won):
• Case 1: \(X = 0\)
• # Explanation:

Step1: Lose or tie both games

The probability of not winning a game in a single - game is \(1 - 0.6=0.4\). Since the games are independent, \(P(X = 0)=(0.3 + 0.1)\times(0.3 + 0.1)=0.4\times0.4 = 0.16\).

• Case 2: \(X = 1\)
• # Explanation:

Step1: Win - lose or win - tie scenarios

The probability of winning the first game and losing the second is \(0.6\times0.3\), the probability of winning the first game and tying the second is \(0.6\times0.1\), the probability of losing the first game and winning the second is \(0.3\times0.6\), and the probability of tying the first game and winning the second is \(0.1\times0.6\).
\(P(X = 1)=0.6\times0.3+0.6\times0.1+0.3\times0.6+0.1\times0.6=2\times(0.6\times0.3)+2\times(0.6\times0.1)=0.36 + 0.12=0.48\).

• Case 3: \(X = 2\)
• # Explanation:

Step1: Win both games

Since the games are independent, \(P(X = 2)=0.6\times0.6 = 0.36\).

The problem involves probability calculations, which fall under the sub - field of Statistics in Mathematics.

For the baseball - team problem:

Step1: Calculate \(P(X = 0)\)

\(P(X = 0)=(1 - 0.45)\times(1 - 0.45)=0.3025\)

Step2: Calculate \(P(X = 1)\)

\(P(X = 1)=2\times0.45\times0.55 = 0.495\)

Step3: Calculate \(P(X = 2)\)

\(P(X = 2)=0.45\times0.45+2\times0.45\times0.55\times0.8=0.2025 + 0.396=0.5985\)

Step1: Calculate \(P(X = 0)\)

\(P(X = 0)=(0.3 + 0.1)\times(0.3 + 0.1)=0.16\)

Step2: Calculate \(P(X = 1)\)

\(P(X = 1)=2\times(0.6\times0.3)+2\times(0.6\times0.1)=0.48\)

Step3: Calculate \(P(X = 2)\)

\(P(X = 2)=0.6\times0.6 = 0.36\)

Answer:

The probability distribution of \(X\) for the baseball team: \(P(X = 0)=0.3025\), \(P(X = 1)=0.495\), \(P(X = 2)=0.5985\)

For the soccer - team problem: