Math Problem Statement

solve question 4 d using simple probability and statistics formula

solve question 4 d using simple probability and statistics formula

Solution

Let's solve Question 4(d) step by step.

The problem is asking us to calculate probabilities related to the number of defective units in a sample of 20, assuming a 100% defective rate in production. This follows a Binomial Distribution where:

  • n=20n = 20 (sample size),
  • p=1p = 1 (probability of a defective unit),
  • XX = the number of defective units.

The binomial probability formula is:

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

where:

  • (nk)\binom{n}{k} = n!k!(n−k)!\frac{n!}{k!(n-k)!} is the number of combinations,
  • pkp^k = probability of kk successes,
  • (1−p)n−k(1-p)^{n-k} = probability of n−kn-k failures.

Since p=1p = 1, the formula simplifies as follows:

\begin{cases} 1 & \text{if } k = 20, \\ 0 & \text{otherwise}. \end{cases}$$ Now let’s solve the subparts. --- ### **(i) Exactly two defective** The probability $$P(X = 2)$$ is calculated as: $$P(X = 2) = \binom{20}{2} \cdot (1)^2 \cdot (0)^{18}.$$ Since $$(1-p) = 0$$, any term where $$X \neq 20$$ has $$P = 0$$: $$P(X = 2) = 0.$$ --- ### **(ii) At most one defective** For this case, $$X = 0$$ or $$X = 1$$: $$P(X \leq 1) = P(X = 0) + P(X = 1).$$ For $$P(X = 0)$$: $$P(X = 0) = \binom{20}{0} \cdot (1)^0 \cdot (0)^{20} = 0.$$ For $$P(X = 1)$$: $$P(X = 1) = \binom{20}{1} \cdot (1)^1 \cdot (0)^{19} = 0.$$ Thus: $$P(X \leq 1) = 0 + 0 = 0.$$ --- ### **(iii) At least two defective** This means $$X \geq 2$$. Since all units are defective ($$p = 1$$), the probability of $$X = 20$$ is: $$P(X = 20) = 1.$$ So: $$P(X \geq 2) = 1.$$ --- ### Final Answers 1. $$P(X = 2) = 0$$, 2. $$P(X \leq 1) = 0$$, 3. $$P(X \geq 2) = 1$$. --- Do you want detailed steps for binomial probability calculations or more clarifications? ### Related Questions 1. How does the binomial distribution change when $$p \neq 1$$? 2. Can you explain why $$P(X = k) = 0$$ for $$k \neq 20$$ in this context? 3. What is the expected value ($$E(X)$$) for this binomial distribution? 4. How would the variance of $$X$$ change if $$p \neq 1$$? 5. What are other real-world examples of this extreme binomial situation? ### Tip: When $$p = 1$$ or $$p = 0$$, binomial calculations become trivial as all probabilities collapse to deterministic outcomes!

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution

Formulas

P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
P(X = k) simplifies when p = 1 or p = 0

Theorems

Binomial Theorem

Suitable Grade Level

Grades 11-12 / Undergraduate

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