Math Problem Statement

Какое наименьшее число снарядов должно быть, чтобы вероятность поражения цели оказалась: а) не ниже чем 0,9; b) не ниже чем 0,99? Вероятность попадания при каждом выстреле равна p = 0,3.

Solution

The problem asks for the minimum number of shots (nn) required for an artillery system to hit a target with a specified probability, given the probability of hitting the target on a single shot (p=0.3p = 0.3). The goal is to ensure that the cumulative probability of hitting the target is at least:

  1. 0.90.9 (part a)
  2. 0.990.99 (part b)

The probability of hitting the target at least once after nn shots is given by the complementary probability:

P(at least one hit)=1P(no hits)=1(1p)nP(\text{at least one hit}) = 1 - P(\text{no hits}) = 1 - (1-p)^n

Calculations

We need to solve the inequality:

1(1p)ntarget probability1 - (1-p)^n \geq \text{target probability}

Rearranging for nn: (1p)n1target probability(1-p)^n \leq 1 - \text{target probability}

Taking the natural logarithm of both sides: nln(1p)ln(1target probability)n \cdot \ln(1-p) \leq \ln(1 - \text{target probability})

Solving for nn: nln(1target probability)ln(1p)n \geq \frac{\ln(1 - \text{target probability})}{\ln(1-p)}

Substituting p=0.3p = 0.3:

  1. For target probability=0.9\text{target probability} = 0.9: nln(10.9)ln(10.3)=ln(0.1)ln(0.7)n \geq \frac{\ln(1 - 0.9)}{\ln(1 - 0.3)} = \frac{\ln(0.1)}{\ln(0.7)}

  2. For target probability=0.99\text{target probability} = 0.99: nln(10.99)ln(10.3)=ln(0.01)ln(0.7)n \geq \frac{\ln(1 - 0.99)}{\ln(1 - 0.3)} = \frac{\ln(0.01)}{\ln(0.7)}

Let me calculate these values. math

Given probability of hitting the target in a single shot

p = 0.3

Target probabilities for parts a and b

target_probabilities = [0.9, 0.99]

Calculate the minimum n for each target probability

results = [] for target_prob in target_probabilities: n = math.ceil(math.log(1 - target_prob) / math.log(1 - p)) results.append(n)

resultsThe results are:

  1. For a target probability of 0.90.9, the minimum number of shots required is n=7n = 7.
  2. For a target probability of 0.990.99, the minimum number of shots required is n=13n = 13.

Let me know if you'd like detailed steps for these calculations or have any other questions!

Expanding on this topic:

  1. How does changing the probability of hitting the target (pp) affect the required number of shots?
  2. How can this formula be applied to other similar scenarios, like reliability in engineering?
  3. What is the effect of rounding on practical implementations of this calculation?
  4. Could you extend this to account for cases with variable probabilities for each shot?
  5. What if the probability pp changes dynamically after each missed shot?

Tip:

The logarithmic formula for cumulative probabilities is widely applicable in reliability analysis, quality control, and survival analysis!

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Math Problem Analysis

Mathematical Concepts

Probability
Exponential Decay
Logarithms

Formulas

P(at least one hit) = 1 - P(no hits) = 1 - (1-p)^n
n ≥ ln(1 - target probability) / ln(1 - p)

Theorems

Logarithmic transformation for inequalities

Suitable Grade Level

High School (Grades 10-12) or Introductory College Level