Question

A baseball player has a 20% chance of hitting one or more home runs in any given game.

1. At the beginning of the season, what is the probability that the player will need six or more games to have his first four games with at least one home run?

2. The player hits at least one home run in three of the first six games of the season. What is the probability that he will have to play two or more additional games to have hos first four games with at least one homerun?

Please answer both parts and clearly box answer

Answer 1

1.

Probability that the player will need six or more games to have his first four games with at least one home run

= Probability that the player won at most three games in first 5 games

Let X be the number of games with at least one home run in 5 games. Then X ~ Binomial(n = 5, p = 0.2)

Probability that the player won at most three games in first 5 games = P(X 3)

= 1 - P(X > 3)

= 1 - [P(X = 4) + P(X = 5)]

= 1 - [⁵C₄ * 0.2⁴ * (1 - 0.2)^5-4  + ⁵C₅ * 0.2⁵ * (1 - 0.2)^5-5 ]

= 1 - (0.00640 + 0.00032)

= 0.99328

2.

Probability that player will have to play two or more additional games to have his first four games with at least one homerun given that he hits at least one home run in three of the first six games

= Probability that player will have to play two or more additional games to have a game with at least one homerun

= Probability that player will not have a game with at least one homerun in a single game

= 1 - 0.2

= 0.8