2.2 Discrete Random Variables
2.2.1 Objectives
By the end of this unit, students will be able to:
- Use probability rules to compute the probability of different types of events.
- Distinguish between marginal probability and conditional probability.
- Apply the multiplication rule to compute probabilities when sampling with/without replacement from finite populations.
- Define random variables.
- Compute expectation and variance of discrete random variables.
2.2.3 Solved Problems
- What is the probability that exactly five of them support the proposition?
Using the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \(n = 10\), \(k = 5\), and \(p = 0.4\).
What is the probability that five or six of them support the proposition?
What is the probability that at least three of them support the proposition?
Exercise: the manufacturer of the ColorSmart-5000 television set claims that 95% of its sets last at least five years without requiring a single repair. Suppose that we contact 8 randomly selected ColorSmart-5000 purchasers five years after they purchased their sets and ask each purchaser: Have you needed any repair for your ColorSmart-5000 TV set during the first 5 years after purchasing the set?
Find the probability that exactly 7 customers needed at least one repair during the first 5 years.
Find the probability that at least 7 purchasers needed at least one repair during the first 5 years.
Solution
Exercise: Profit from crop yield under different weather condi-tions (X). 1. Determine the missing probability in the following distribution.
| Weather | Profit ($) | Probability |
|---|---|---|
| Dry | 200,000 | 0.30 |
| Light Rain | 300,000 | 0.50 |
| Storm | 150,000 | ? |
- Find the expected profit from this crop, \(\mu\).
\[ \begin{aligned} \mu &= \sum{x_i}P(x_i) \\ &= \$200k \cdot 0.3 + \$300k \cdot 0.5 + \$150k \cdot 0.2 \\ &= \$60k + \$150k + \$30k = \$240k \end{aligned} \]
- Find the variance, \(\sigma^2\) and the standard deviation, \(\sigma\).
\[ \begin{aligned} \sigma^2 &= \sum(x_i-\mu)^2P(x_i) \\ &= (\$200k-\$240k)^2 \ . 0.3 + (\$300k-\$240k)^2 \ . 0.5 + (\$150k-\$240)^2 . 0.2 \\ &= \$^2 3,900, 000, 000 \\ \end{aligned} \]
The standard deviation is \(\sigma = \sqrt{\$^2 3,900,000,000}\) = $62,450
- Interpret the value of the expected profit, \(\mu\).
The expected profit represents the long-run average, the expected profit on average in the future.
Example/Exercise: 40% of all voters support Proposition A. If a random sample of 10 voters is polled. Find the following probabilities.
- What is the probability that exactly five of them support the proposition?
\[ \begin{aligned} P(X=5)&=\frac{10!}{(10-5)!5!}(0.40)^5(0.60)^{10-5} \\ &=\frac{10!}{5!5!}(0.40)^5(0.60)^5 \\ &=(252)(0.01024)(0.07776) \\ &=0.2007 \end{aligned} \]
- What is the probability that five or six of them support the proposition?
\[ \begin{aligned} P(X=5)+P(X=6)&=0.2007+\frac{10!}{(10-6)!6!}(0.40)^6(0.60)^{10-6} \\ &=0.2007+\frac{10!}{4!6!}(0.40)^6(0.60)^4 \\ &=0.2007+(210)(0.004096)(0.1296) \\ &=0.2007+0.1115 \\ &=0.312 \end{aligned} \]
- What is the probability that at least three of them support the proposition?
\[ \begin{aligned} P(x \ge 3) &= P(X=3)+P(X=4)+P(X=5)+P(X=6)+...+P(X=10) \\ &=1-[P(X=0)+P(X=1)+P(X=2)] \\ &=1-[\frac{10!}{0!10!}(0.40)^0(0.60)^{10}+\frac{10!}{1!9!}(0.40)^1(0.60)^9+\frac{10!}{2!8!}(0.40)^2(0.60)^8] \\ &=1-[(1)(1)(0.0060)+(10)(0.40)(0.0101)+(45)(0.16)(0.0168)] \\ &=1-[0.0060+0.0403+0.1209] \\ &=1-0.1672 = 0.8328 \end{aligned} \]
Exercise: the manufacturer of the ColorSmart-5000 television set claims that 95% of its sets last at least five years without requiring a single repair. Suppose that we contact 8 randomly selected ColorSmart-5000 purchasers five years after they purchased their sets and ask each purchaser: Have you needed any repair for your ColorSmart-5000 TV set during the first 5 years after purchasing the set?
- Find the probability that exactly 7 customers needed at least one repair during the first 5years.
\[ \begin{aligned} P(X=7)&=\frac{8!}{(8-7)!7!}(0.05)^7(0.95)^{8-7} \\ &=\frac{8!}{1!7!}(0.05)^7(0.95)^1 \\ &=0.0000000059375 \end{aligned} \]
- Find the probability that at least 7 purchasers needed at least one repair during the first 5years.
\[ \begin{aligned} P(X=7)&=P(X=7)+P(X=8)\\ &=0.0000000059375+\frac{8!}{(8-8)!8!}(0.05)^8(0.95)^{8-8} \\ &=0.0000000059375+0.0000000000390625 \\ &=0.0000000059765625 \end{aligned} \]
2.2.4 Exercises
Exercise: Profit from crop yield under different weather conditions (X).
- Determine the missing probability in the following distribution.
| Weather | Profit ($) | Probability |
|---|---|---|
| Dry | 200,000 | 0.30 |
| Light Rain | 300,000 | 0.50 |
| Storm | 150,000 | ? |
- Find the expected profit from this crop, \(\mu\).
| \(x_i\) | \(P(x_i)\) | \(x_i P(x_i)\) |
|---|---|---|
| 200,000 | 0.30 | \(200,000 \times 0.30\) |
| 300,000 | 0.50 | \(300,000 \times 0.50\) |
| 150,000 | ? | \(150,000 \times ?\) |
- Find the variance, \(\sigma^2\) and the standard deviation, \(\sigma\).
| \(x_i\) | \(P(x_i)\) | \(\mu\) | \((x_i - \mu)\) | \((x_i - \mu)^2\) | \((x_i - \mu)^2 P(x_i)\) |
|---|---|---|---|---|---|
| 200,000 | 0.30 | ||||
| 300,000 | 0.50 | ||||
| 150,000 | ? |
The standard deviation is \(\sigma\) = ________________
- Interpret the value of the expected profit, \(\mu\).
Example/Exercise: 40% of all voters support Proposition A. If a random sample of 10 voters is polled. Find the following probabilities.
Formula for the probability of exactly \(x\) successes from \(n\) trials
\[ p(x) = \binom{n}{x} p^x q^{n-x} = \frac{n!}{(n-x)!x!} p^x q^{n-x}; \quad \text{where } x = 0, 1, 2, \ldots, n \]
and
\[ n! = n(n-1)(n-2) \cdots (3)(2)(1) \]