Bernoulli Trials and Binomial Distribution

Bernoulli Trials

Many experiments have outcomes that can be classified into two categories: Success or Failure. For example, tossing a coin (Head or Tail), a manufactured item being Defective or Non-defective, or an examinee passing or failing.

Trials of a random experiment are called Bernoulli trials if they satisfy the following four conditions:

1. The number of trials is finite.
2. The trials are independent of each other.
3. Each trial has exactly two outcomes: success (S) or failure (F).
4. The probability of success ($p$) remains the same in each trial.

If $p$ is the probability of success, then the probability of failure is $$q = 1 - p$$

Binomial Distribution

Consider a series of $n$ Bernoulli trials. Let $X$ be a random variable representing the number of successes in these $n$ trials. Then $X$ is said to follow a Binomial Distribution with parameters $n$ and $p$.

The Probability Mass Function (PMF): The probability of getting exactly $r$ successes in $n$ trials is $$P(X = r) = \binom{n}{r} p^r q^{n-r}$$ where $$r = 0,1,2,\dots,n$$

Notation: $$X \sim B(n,p)$$

Key Parameters:

• Mean (Expected Value): $$E(X) = np$$
• Variance: $$Var(X) = npq$$
• Standard Deviation: $$\sigma = \sqrt{npq}$$

Example 1: Basic Coin Toss

If a fair coin is tossed 10 times, find the probability of getting exactly 6 heads.

Solution: Let $X$ be the number of heads obtained.

For a fair coin, $$n = 10, \quad p = \frac{1}{2}, \quad q = \frac{1}{2}$$ We need $$P(X=6)$$ Using the binomial formula, $$P(X=6)=\binom{10}{6}\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^4$$ $$=\binom{10}{6}\left(\frac{1}{2}\right)^{10}$$ Now, $$\binom{10}{6}=\binom{10}{4}=\frac{10\cdot 9\cdot 8\cdot 7}{4\cdot 3\cdot 2\cdot 1}=210$$ Therefore, $$P(X=6)=210\cdot \frac{1}{1024}=\frac{210}{1024}=\frac{105}{512}$$

Answer: $$\boxed{\frac{105}{512}}$$

Example 2: Dice Rolling - Multiple Successes

A die is thrown 6 times. If 'getting an odd number' is a success, find the probability of at least 5 successes.

Solution: A success means getting one of $$\{1,3,5\}$$ Hence, $$p = \frac{3}{6}=\frac{1}{2}, \quad q=\frac{1}{2}, \quad n=6$$

We need $$P(X\ge 5)=P(X=5)+P(X=6)$$ Now, $$P(X=5)=\binom{6}{5}\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^1$$ $$=6\left(\frac{1}{2}\right)^6=\frac{6}{64}$$ Also, $$P(X=6)=\binom{6}{6}\left(\frac{1}{2}\right)^6=\frac{1}{64}$$ Therefore, $$P(X\ge 5)=\frac{6}{64}+\frac{1}{64}=\frac{7}{64}$$

Answer: $$\boxed{\frac{7}{64}}$$

Example 3: Finding Mean and Variance

A binomial distribution has $n = 100$ and $p = 0.2$. Find its mean and standard deviation.

Solution: Given, $$n=100, \quad p=0.2$$ Then, $$q=1-p=0.8$$

Mean is $$E(X)=np=100\times 0.2=20$$ Variance is $$Var(X)=npq=100\times 0.2\times 0.8=16$$ Standard deviation is $$\sigma=\sqrt{16}=4$$

Answer: $$\boxed{\text{Mean}=20, \quad \text{S.D.}=4}$$

Example 4: Target Hitting

The probability of a man hitting a target is $\frac{1}{4}$. If he fires 7 times, what is the probability of hitting the target at least once?

Solution: Let success be “hitting the target”. Then, $$n=7, \quad p=\frac{1}{4}, \quad q=\frac{3}{4}$$ We need $$P(X\ge 1)$$ Using the complement rule, $$P(X\ge 1)=1-P(X=0)$$ Now, $$P(X=0)=\binom{7}{0}\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^7=\left(\frac{3}{4}\right)^7$$ $$=\frac{2187}{16384}$$ Hence, $$P(X\ge 1)=1-\frac{2187}{16384}$$ $$=\frac{16384-2187}{16384}=\frac{14197}{16384}$$

Answer: $$\boxed{\frac{14197}{16384}}$$

Example 5: Finding $n$ from Mean and Variance

For a binomial distribution, the mean is 6 and the variance is 2. Find the parameters $n$ and $p$.

Solution: For a binomial distribution, $$np=6 \qquad \text{and} \qquad npq=2$$ Divide the second equation by the first: $$\frac{npq}{np}=\frac{2}{6}$$ $$q=\frac{1}{3}$$ Therefore, $$p=1-q=1-\frac{1}{3}=\frac{2}{3}$$ Now substitute into $$np=6$$ So, $$n\cdot \frac{2}{3}=6$$ $$2n=18$$ $$n=9$$

Answer: $$\boxed{n=9, \quad p=\frac{2}{3}}$$

Example 6: Defective Items - At Most

In a large consignment of electric bulbs, 10% are defective. A random sample of 20 bulbs is drawn. Find the probability that at most 2 bulbs are defective.

Solution: Let $X$ denote the number of defective bulbs in the sample. Then, $$n=20, \quad p=0.1, \quad q=0.9$$ We need $$P(X\le 2)=P(X=0)+P(X=1)+P(X=2)$$ Now, $$P(X=0)=\binom{20}{0}(0.1)^0(0.9)^{20}=(0.9)^{20}$$ $$P(X=1)=\binom{20}{1}(0.1)^1(0.9)^{19}=20(0.1)(0.9)^{19}=2(0.9)^{19}$$ $$P(X=2)=\binom{20}{2}(0.1)^2(0.9)^{18}=190(0.01)(0.9)^{18}=1.9(0.9)^{18}$$ Hence, $$P(X\le 2)=(0.9)^{20}+2(0.9)^{19}+1.9(0.9)^{18}$$ Taking $(0.9)^{18}$ common, $$P(X\le 2)=(0.9)^{18}\big[(0.9)^2+2(0.9)+1.9\big]$$ $$=(0.9)^{18}(0.81+1.8+1.9)$$ $$=(0.9)^{18}(4.51)$$

Answer: $$\boxed{P(X\le 2)=4.51(0.9)^{18}}$$

Example 7: Ratio of Probabilities

In a binomial distribution $B(n,p)$, if $P(X=1)=P(X=2)$, find the relationship between $n$ and $p$.

Solution: Using the binomial formula, $$P(X=1)=\binom{n}{1}p q^{n-1}$$ $$P(X=2)=\binom{n}{2}p^2 q^{n-2}$$ Given, $$\binom{n}{1}p q^{n-1}=\binom{n}{2}p^2 q^{n-2}$$ Substituting the combinations, $$n p q^{n-1}=\frac{n(n-1)}{2}p^2 q^{n-2}$$ Cancelling common factors $n$, $p$, and $q^{n-2}$, $$q=\frac{n-1}{2}p$$ Since $$q=1-p$$ we get $$1-p=\frac{n-1}{2}p$$ Multiply by 2: $$2-2p=(n-1)p$$ $$2=(n+1)p$$ Hence, $$p=\frac{2}{n+1}$$

Answer: $$\boxed{p=\frac{2}{n+1}}$$

Example 8: Multiple Choice Guessing

A student answers 5 multiple choice questions by guessing. Each question has 4 options (one correct). Find the probability that he gets at least 4 correct.

Solution: For each question, $$p=\frac{1}{4}, \quad q=\frac{3}{4}, \quad n=5$$ We need $$P(X\ge 4)=P(X=4)+P(X=5)$$ Now, $$P(X=4)=\binom{5}{4}\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right)$$ $$=5\cdot \frac{1}{256}\cdot \frac{3}{4}=\frac{15}{1024}$$ Also, $$P(X=5)=\binom{5}{5}\left(\frac{1}{4}\right)^5=\frac{1}{1024}$$ Therefore, $$P(X\ge 4)=\frac{15}{1024}+\frac{1}{1024}=\frac{16}{1024}=\frac{1}{64}$$

Answer: $$\boxed{\frac{1}{64}}$$

Example 9: Sum of Mean and Variance

For a binomial distribution $B(n,p)$, the sum of mean and variance is 15 and the product is 54. Find $n$.

Solution: For a binomial distribution, $$\text{Mean}=np, \qquad \text{Variance}=npq$$ Given, $$np+npq=15$$ $$np(1+q)=15 \qquad \text{...(1)}$$ Also, $$(np)(npq)=54$$ $$(np)^2 q=54 \qquad \text{...(2)}$$ Let $$np=k$$ Then from (1), $$k(1+q)=15$$ $$k=\frac{15}{1+q}$$ Substitute into (2): $$\left(\frac{15}{1+q}\right)^2 q=54$$ $$\frac{225q}{(1+q)^2}=54$$ $$225q=54(1+2q+q^2)$$ $$225q=54+108q+54q^2$$ $$54q^2-117q+54=0$$ Divide by 9: $$6q^2-13q+6=0$$ Factorize: $$(3q-2)(2q-3)=0$$ So, $$q=\frac{2}{3} \quad \text{or} \quad q=\frac{3}{2}$$ But probability cannot exceed 1, hence $$q=\frac{2}{3}$$ Therefore, $$p=1-q=\frac{1}{3}$$ Now from (1), $$np(1+q)=15$$ $$np\left(1+\frac{2}{3}\right)=15$$ $$np\cdot \frac{5}{3}=15$$ $$np=9$$ Since $$p=\frac{1}{3}$$ we get $$n\cdot \frac{1}{3}=9$$ $$n=27$$

Answer: $$\boxed{n=27}$$

Example 10: Pair of Dice - Success on Sum

A pair of dice is thrown 4 times. If getting a sum of 9 is considered a success, find the probability of exactly 2 successes.

Solution: A success means obtaining sum 9. The favorable outcomes are $$(3,6), (4,5), (5,4), (6,3)$$ So for one trial, $$p=\frac{4}{36}=\frac{1}{9}, \quad q=\frac{8}{9}, \quad n=4$$ We need $$P(X=2)$$ Using the binomial formula, $$P(X=2)=\binom{4}{2}\left(\frac{1}{9}\right)^2\left(\frac{8}{9}\right)^2$$ $$=6\cdot \frac{1}{81}\cdot \frac{64}{81}$$ $$=\frac{384}{6561}$$ Now simplify: $$\frac{384}{6561}=\frac{128}{2187}$$

Answer: $$\boxed{\frac{128}{2187}}$$