Math Problem Statement

A high school baseball player has a 0.182 batting average. In one game, he gets 5 at bats. What is the probability he will get at least 2 hits in the game?

Solution

To solve this, we need to use the binomial probability distribution since the number of hits in the game follows a binomial pattern: each at-bat is an independent trial with two outcomes (hit or no hit). Here's how we proceed:

Problem Breakdown:

  • The player's batting average (probability of getting a hit in a given at-bat) is p=0.182p = 0.182.
  • The number of at-bats (trials) in the game is n=5n = 5.
  • We need to find the probability that the player gets at least 2 hits in 5 at-bats, which means the player gets 2 or more hits.

Let XX be the random variable representing the number of hits the player gets. We are looking for:

P(X2)P(X \geq 2)

This is equivalent to:

P(X2)=1P(X<2)=1(P(X=0)+P(X=1))P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1))

We'll now calculate P(X=0)P(X = 0) and P(X=1)P(X = 1) using the binomial probability formula:

P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}

Where (nk)\binom{n}{k} is the binomial coefficient and represents the number of ways to choose kk successes (hits) from nn trials (at-bats).

Step 1: Compute P(X=0)P(X = 0)

P(X=0)=(50)(0.182)0(10.182)5=1×1×(0.818)5P(X = 0) = \binom{5}{0} (0.182)^0 (1 - 0.182)^5 = 1 \times 1 \times (0.818)^5

P(X=0)=(0.818)50.340P(X = 0) = (0.818)^5 \approx 0.340

Step 2: Compute P(X=1)P(X = 1)

P(X=1)=(51)(0.182)1(10.182)4=5×0.182×(0.818)4P(X = 1) = \binom{5}{1} (0.182)^1 (1 - 0.182)^4 = 5 \times 0.182 \times (0.818)^4

P(X=1)=5×0.182×0.3790.345P(X = 1) = 5 \times 0.182 \times 0.379 \approx 0.345

Step 3: Compute P(X2)P(X \geq 2)

Now, we can calculate the probability of getting at least 2 hits:

P(X2)=1(P(X=0)+P(X=1))=1(0.340+0.345)=10.685=0.315P(X \geq 2) = 1 - (P(X = 0) + P(X = 1)) = 1 - (0.340 + 0.345) = 1 - 0.685 = 0.315

Final Answer:

The probability that the player will get at least 2 hits in the game is approximately 0.315 or 31.5%.


Would you like more details on any part of the calculation, or do you have other questions? Here are some related topics you might find interesting:

  1. What is the binomial probability distribution and when do we use it?
  2. How do we calculate binomial coefficients?
  3. What are the key properties of a binomial distribution?
  4. How does changing the probability of success affect the outcome of a binomial distribution?
  5. Can we use this method to calculate the probability of exactly 3 hits or more?

Tip: When working with binomial probabilities, always remember to check if the trials are independent and if the probability of success remains constant for each trial.

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Math Problem Analysis

Mathematical Concepts

Binomial Probability Distribution
Probability Theory
Combinatorics

Formulas

P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
P(X ≥ k) = 1 - P(X < k) = 1 - (P(X = 0) + P(X = 1))

Theorems

Binomial Theorem

Suitable Grade Level

Grades 9-12