Discrete Probability Distributions

A Bernoulli process must satisfy the following conditions:

1. The experiment consists of repeated trials.

2. Each trial results in an outcome that may be classified as a success or a failure.

3. The probability of success, denoted by \( p \), remains constant from trial to trial.

4. The repeated trials are independent.

A Bernoulli trial results in:

• Success with probability \( p \)

• Failure with probability \( q = 1 - p \)

The probability distribution of the Binomial random variable \( X \), representing the number of successes in \( n \) independent trials, is given by:

\[ b(x; n, p) = \binom{n}{x} p^x (1 - p)^{n - x}, \quad \text{for } x = 0, 1, 2, \ldots, n \]

The mean and variance of the binomial distribution \( b(x; n, p) \) are given by: \[ \mu = np \quad \text{and} \quad \sigma^2 = npq \] where \( q = 1 - p \).

A probability distribution problem is likely a Binomial Distribution if the following conditions (and keywords) are present:

1. Fixed number of trials ($n$): Look for phrases like:

   • “out of 10 attempts”, “for 20 customers”, “tested 5 items”

2. Each trial has two possible outcomes (Success/Failure): Keywords may include:

   • “defective/non-defective”, “pass/fail”, “survive/die”, “hit/miss”

3. Constant probability of success ($p$) in each trial: Phrases like:

   • “with probability 0.3”, “each has a 60% chance”

4. Independent trials: Often stated or implied as:

   • “each trial is independent”, “randomly chosen”

5. Questions asking “exactly $x$ successes”, “at least $x$ successes”, or “between $a$ and $b$ successes” often indicate binomial distribution.

Hint: If you're counting the number of times something happens in a fixed number of identical and independent trials with a fixed success probability, you're likely dealing with a binomial distribution.

A fair coin is tossed 5 times. What is the probability of getting exactly 3 heads? Also compute the mean and variance of the distribution.

Solution:
Here, \( n = 5, p = 0.5, q = 0.5 \)
Let \( X \) be the number of heads. Then, \[ P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^2 = 10 \times 0.125 \times 0.25 = 0.3125 \] Mean: \( \mu = np = 5 \times 0.5 = 2.5 \)
Variance: \( \sigma^2 = npq = 5 \times 0.5 \times 0.5 = 1.25 \)

A machine produces bulbs with a 10% defect rate. If 8 bulbs are chosen at random, what is the probability that exactly 2 are defective? Also find the mean and variance.

Solution:
Here, \( n = 8, p = 0.1, q = 0.9 \) \[ P(X = 2) = \binom{8}{2} (0.1)^2 (0.9)^6 = 28 \times 0.01 \times 0.531441 = 0.1488 \] Mean: \( \mu = np = 8 \times 0.1 = 0.8 \)
Variance: \( \sigma^2 = npq = 8 \times 0.1 \times 0.9 = 0.72 \)

A student guesses on a multiple choice question with 4 options. If there are 10 questions, what is the probability of getting exactly 4 correct answers by guessing?

Solution:
Here, \( n = 10, p = \frac{1}{4} = 0.25, q = 0.75 \) \[ P(X = 4) = \binom{10}{4} (0.25)^4 (0.75)^6 = 210 \times 0.00390625 \times 0.1779785 \approx 0.146 \] Mean: \( \mu = np = 10 \times 0.25 = 2.5 \)
Variance: \( \sigma^2 = npq = 10 \times 0.25 \times 0.75 = 1.875 \)

The probability of a student passing an exam is 0.6. If 7 students appear for the exam, find the probability that exactly 5 pass. Also calculate mean and variance.

Solution:
Here, \( n = 7, p = 0.6, q = 0.4 \) \[ P(X = 5) = \binom{7}{5} (0.6)^5 (0.4)^2 = 21 \times 0.07776 \times 0.16 = 0.2613 \] Mean: \( \mu = np = 7 \times 0.6 = 4.2 \)
Variance: \( \sigma^2 = npq = 7 \times 0.6 \times 0.4 = 1.68 \)

In a quality control check, the probability that an item is of good quality is 0.85. If 12 items are tested, find the probability that exactly 10 are of good quality. Also find the mean and variance.

Solution:
Here, \( n = 12, p = 0.85, q = 0.15 \) \[ P(X = 10) = \binom{12}{10} (0.85)^{10} (0.15)^2 = 66 \times 0.19687 \times 0.0225 \approx 0.292 \] Mean: \( \mu = np = 12 \times 0.85 = 10.2 \)
Variance: \( \sigma^2 = npq = 12 \times 0.85 \times 0.15 = 1.53 \)

The probability that a student passes a final exam is $0.6$. If $12$ students take the exam, what is the probability that

1. [(a)] at least $8$ pass,

2. [(b)] between $4$ and $9$ pass, and

3. [(c)] exactly $6$ pass?

Solution:
Let $X$ be the number of students who pass. Then $X$ follows a binomial distribution with parameters $n = 12$, $p = 0.6$.

(a) Probability that at least 8 pass: \[ P(X \geq 8) = 1 - P(X \leq 7) = 1 - \sum_{x=0}^{7} b(x; 12, 0.6) \] Using cumulative binomial values: \[ P(X \geq 8) = 1 - 0.7831 = \boxed{0.2169} \]

(b) Probability that between 4 and 9 pass: \[ P(4 \leq X \leq 9) = \sum_{x=4}^{9} b(x; 12, 0.6) = \sum_{x=0}^{9} b(x; 12, 0.6) - \sum_{x=0}^{3} b(x; 12, 0.6) \] \[ P(4 \leq X \leq 9) = 0.9450 - 0.0635 = \boxed{0.8815} \]

(c) Probability that exactly 6 pass: \[ P(X = 6) = b(6; 12, 0.6) = \sum_{x=0}^{6} b(x; 12, 0.6) - \sum_{x=0}^{5} b(x; 12, 0.6) \] \[ P(X = 6) = 0.6100 - 0.4070 = \boxed{0.2030} \]

A discrete uniform distribution has a finite number of outcomes that are equally likely.

PMF: \[ P(X = x) = \frac{1}{n}, \quad x = x_1, x_2, \dots, x_n \]

Number of outcomes: \[ n = \frac{b-a}{d} + 1, \quad d = \text{spacing between consecutive numbers} \]

Mean: \[ E[X] = \frac{a+b}{2} \]

Variance: \[ \mathrm{Var}(X) = \frac{(n^2 - 1)d^2}{12} \]

• Problem states all outcomes are "equally likely"

• Finite number of possible outcomes

• Often includes phrases like:

  • "Choose a number at random from 1 to n"

  • "Fair die / coin / card deck"

  • "Equally probable outcomes"

Problem: A fair 12-sided die is rolled. Compute the expected value, variance, and the probability that the outcome is greater than 8.

Solution:

• Here $a=1, b=12, d=1, n=12$ (numbers 1 to 12)

• Mean: $E[X] = \frac{a+b}{2} = \frac{1+12}{2} = 6.5$

• Variance: $\text{Var}(X) = \frac{(n^2-1)d^2}{12} = \frac{(12^2-1)\cdot1^2}{12} = \frac{143}{12} \approx 11.92$

• Probability $X>8$: favorable outcomes $\{9,10,11,12\} \Rightarrow 4$ out of 12 \[ P(X>8) = \frac{4}{12} = \frac{1}{3} \approx 0.333 \]

Problem: A number is chosen randomly from integers 10 to 25. Compute the expected value, variance, and the probability that the number is $\le 15$.

Solution:

• $a=10, b=25, d=1, n = b-a+1 = 16$

• Mean: $E[X] = \frac{a+b}{2} = \frac{10+25}{2} = 17.5$

• Variance: $\text{Var}(X) = \frac{(n^2-1)d^2}{12} = \frac{(16^2-1)\cdot1^2}{12} = \frac{255}{12} = 21.25$

• Probability $X\le 15$: numbers 10 to 15 $\Rightarrow 6$ out of 16 \[ P(X\le15) = \frac{6}{16} = \frac{3}{8} = 0.375 \]

Problem: A deck contains 6 cards numbered 1 to 6. Compute the expected value, variance, and probability that the card number is $>4$.

Solution:

• $a=1, b=6, d=1, n=6$

• Mean: $E[X] = \frac{1+6}{2} = 3.5$

• Variance: $\text{Var}(X) = \frac{(6^2-1)\cdot1^2}{12} = \frac{35}{12} \approx 2.92$

• Probability $X>4$: favorable outcomes $\{5,6\} \Rightarrow 2/6$ \[ P(X>4) = \frac{2}{6} = \frac{1}{3} \approx 0.333 \]

Problem: A raffle ticket is randomly selected from tickets numbered 20 to 60. Compute expected value, variance, and probability ticket $\ge 50$.

Solution:

• $a=20, b=60, d=1, n = b-a+1 = 41$

• Mean: $E[X] = \frac{20+60}{2} = 40$

• Variance: $\text{Var}(X) = \frac{(41^2-1)\cdot1^2}{12} = \frac{1680}{12} = 140$

• Probability $X\ge50$: numbers 50 to 60 $\Rightarrow 11/41$ \[ P(X\ge50) = \frac{11}{41} \approx 0.268 \]

Problem: A number is picked randomly from \(\{2,4,6,8,10,12\}\). Compute expected value, variance, probability $<7$.

Solution:

• $a=2, b=12, d=2, n = \frac{b-a}{d}+1 = \frac{12-2}{2}+1 = 6$

• Mean: $E[X] = \frac{2+12}{2} = 7$

• Variance: $\text{Var}(X) = \frac{(6^2-1)\cdot2^2}{12} = \frac{35\cdot4}{12} \approx 11.667$

• Probability $X<7$: numbers 2,4,6 $\Rightarrow 3/6$ \[ P(X<7) = \frac{3}{6} = 0.5 \]