Math Problem Statement

Вероятность попадания по мишени при каждом выстреле некоторым стрелком равна 0.9. Найди вероятность попадания по мишени в серии из семи выстрелов этим стрелком: 1) по крайней мере 2 раза; 2) по крайней мере 1 раз.

Solution

The task involves calculating probabilities in a binomial distribution setting. Here's a detailed breakdown of the problem:

Given:

  • Probability of hitting the target in one shot: p=0.9p = 0.9.
  • Number of shots: n=7n = 7.
  • Complement probability (miss): q=1p=0.1q = 1 - p = 0.1.

You are asked to calculate the following:

  1. The probability of hitting the target at least 2 times in 7 shots.
  2. The probability of hitting the target at least 1 time in 7 shots.

Solution:

The binomial probability formula is: P(X=k)=C(n,k)pkqnk,P(X = k) = C(n, k) \cdot p^k \cdot q^{n-k}, where:

  • C(n,k)=n!k!(nk)!C(n, k) = \frac{n!}{k!(n-k)!} is the number of combinations,
  • pp is the probability of success,
  • q=1pq = 1-p is the probability of failure,
  • kk is the number of successes.

To compute "at least" probabilities, we use: P(Xk)=1P(X<k)=1i=0k1P(X=i).P(X \geq k) = 1 - P(X < k) = 1 - \sum_{i=0}^{k-1} P(X = i).


1) Probability of hitting at least 2 times:

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

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

P(X=0)=C(7,0)(0.9)0(0.1)7=110.0000001=0.0000001.P(X = 0) = C(7, 0) \cdot (0.9)^0 \cdot (0.1)^7 = 1 \cdot 1 \cdot 0.0000001 = 0.0000001.

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

P(X=1)=C(7,1)(0.9)1(0.1)6=70.90.000001=0.0000063.P(X = 1) = C(7, 1) \cdot (0.9)^1 \cdot (0.1)^6 = 7 \cdot 0.9 \cdot 0.000001 = 0.0000063.

Step 3: Subtract:

P(X2)=1P(X=0)P(X=1)=10.00000010.00000630.9999936.P(X \geq 2) = 1 - P(X = 0) - P(X = 1) = 1 - 0.0000001 - 0.0000063 \approx 0.9999936.


2) Probability of hitting at least 1 time:

P(X1)=1P(X=0).P(X \geq 1) = 1 - P(X = 0).

Step 1: Use P(X=0)P(X = 0) from above:

P(X1)=10.0000001=0.9999999.P(X \geq 1) = 1 - 0.0000001 = 0.9999999.


Final Answers:

  1. P(X2)0.9999936P(X \geq 2) \approx 0.9999936,
  2. P(X1)0.9999999P(X \geq 1) \approx 0.9999999.

Would you like me to provide step-by-step calculations or visualizations for binomial probabilities?


Related Questions:

  1. What is the probability of hitting the target exactly 3 times in 7 shots?
  2. How do you calculate P(X=k)P(X = k) for any given kk in a binomial distribution?
  3. What are the expected value and variance of this binomial experiment?
  4. How do probabilities change if pp is reduced to 0.8 instead of 0.9?
  5. How does the complement rule simplify probability calculations?

Tip:

When solving binomial problems, knowing the cumulative probabilities (using tables or software) can save significant time for larger values of nn.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution
Complement Rule

Formulas

P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
P(X >= k) = 1 - P(X < k)

Theorems

Binomial Probability Formula

Suitable Grade Level

Grades 10-12