Math Problem Statement

A clay pigeon shooter hits a clay with probability 0.65. There are 25 clays in each set. To win a set, the shooter must hit at least 15 clays. The shooter takes part in 10 sets of 25 clays. Find the probability that the shooter wins: (a) exactly 7 of these sets, (b) fewer than 5 of these sets.

Solution

The problem provided involves calculating probabilities related to a clay pigeon shooter using binomial distribution concepts. Let's break down the solution step by step.

Problem Details:

  • The shooter hits a clay with a probability of p=0.65p = 0.65.
  • There are 25 clays in each set, and to win a set, the shooter must hit at least 15 clays.
  • The shooter participates in 10 sets of 25 clays.

Tasks:

  1. Find the probability that the shooter wins exactly 7 of these 10 sets.
  2. Find the probability that the shooter wins fewer than 5 of these sets.

Step 1: Determine the probability of winning a single set

To find the probability of the shooter winning a set (hitting at least 15 out of 25 clays), we use the binomial distribution:

  • Let XX be the number of clays hit in a single set. XX follows a binomial distribution: XBinomial(n=25,p=0.65)X \sim \text{Binomial}(n=25, p=0.65)
  • The shooter wins a set if X15X \geq 15. Therefore, we need to calculate: P(Win)=P(X15)=k=1525(25k)(0.65)k(0.35)25kP(\text{Win}) = P(X \geq 15) = \sum_{k=15}^{25} \binom{25}{k} (0.65)^k (0.35)^{25-k}

Step 2: Use this probability in a binomial distribution for multiple sets

Let YY be the number of sets won out of 10. YY also follows a binomial distribution: YBinomial(n=10,p=P(Win))Y \sim \text{Binomial}(n=10, p=P(\text{Win}))

Now, we can calculate the required probabilities.

(a) Probability of winning exactly 7 sets

P(Y=7)=(107)(P(Win))7(1P(Win))3P(Y = 7) = \binom{10}{7} (P(\text{Win}))^7 (1-P(\text{Win}))^3

(b) Probability of winning fewer than 5 sets

P(Y<5)=P(Y=0)+P(Y=1)+P(Y=2)+P(Y=3)+P(Y=4)P(Y < 5) = P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3) + P(Y = 4)

Calculations:

I'll calculate the exact values for these probabilities.


Let's calculate these probabilities using Python.### Solution:

  1. Probability of winning a single set: P(Win)0.771P(\text{Win}) \approx 0.771.

  2. Probability of winning exactly 7 out of 10 sets: P(Y=7)0.233P(Y = 7) \approx 0.233

  3. Probability of winning fewer than 5 out of 10 sets: P(Y<5)0.0127P(Y < 5) \approx 0.0127

These values are the probabilities for the two scenarios given in the problem.

Summary:

  • (a) The probability that the shooter wins exactly 7 out of 10 sets is approximately 0.233.
  • (b) The probability that the shooter wins fewer than 5 out of 10 sets is approximately 0.0127.

Would you like more details or have any questions?


Related Questions:

  1. How does the probability change if the shooter's accuracy improves to 0.75?
  2. What is the expected number of sets won by the shooter out of 10?
  3. How would the probabilities change if the shooter needed to hit at least 20 clays to win a set?
  4. Can the normal approximation be used to estimate the probability of winning at least 7 sets? If so, how accurate is it?
  5. How does the probability of winning a set change if the number of clays per set increases to 30?

Tip: When dealing with multiple events, the binomial distribution is a powerful tool to calculate the likelihood of a specific number of successes, given a known success probability for each individual event.

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Math Problem Analysis

Mathematical Concepts

Binomial Distribution
Probability Theory
Combinatorics

Formulas

P(Win) = P(X ≥ 15) = Σ from k=15 to 25 [C(25, k) * (0.65)^k * (0.35)^(25-k)]
Y ~ Binomial(n=10, p=P(Win))
P(Y = 7) = C(10, 7) * (P(Win))^7 * (1-P(Win))^3
P(Y < 5) = P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3) + P(Y = 4)

Theorems

Binomial Theorem
Law of Total Probability

Suitable Grade Level

Grade 12 / Early College