Question

suppose you attend a baseball game late in the season. The Houston Astros are playing, and jeff bagwells batting average is .308 what is the probability he will get his first hit of the game on or before his third time at bat. a)0.669 b)0.147 c)0.924 d) 0.029 or e)0.308

Answer 1

Using Jeff Bagwell's batting average of .308, the probability that he gets at least one hit on or before his third time at bat is calculated as 1 minus the probability of not getting any hits in three at-bats, which is approximately 0.669.

Therefore, the correct answer is: option a) 0.669

Step-by-step explanation:

To calculate the probability that Jeff Bagwell will get his first hit of the game on or before his third time at bat, we use his batting average which is .308.

This average means he has a 30.8% chance of getting a hit each time he is at bat. Using the complementary probability, we determine the chances that he does not get a hit during his at-bats, and then subtract from 1 to find the probability of at least one hit in three at-bats.

The probability of not getting a hit in one at-bat is 1 - 0.308 which is 0.692.

So, the probability of not getting a hit in three consecutive at-bats would be 0.692 raised to the power of three.

Now, let's calculate the probability of not getting any hits:

P(No hits in three at-bats) = 0.692^3

P(No hits in three at-bats) ≈ 0.331.

Therefore, the probability he gets at least one hit in his first three at-bats is:

P(At least one hit in three at-bats) = 1 - P(No hits in three at-bats)

P(At least one hit in three at-bats) ≈ 1 - 0.331

P(At least one hit in three at-bats) ≈ 0.669.