Probability Distributions

This post is about some solved Binomial distribution Questions. These solved binomial distribution questions make use of computation of (i) the exact probability case, (ii) at least case, (iii) at most case, and (iv) between cases.

• The sum of all probabilities in the distribution sums up to 1
• The probability of success in all $n$ trials is $p^n$
• The probability of failure in all $n$ trials is $(1 – p)^n = q^n$
• Probability of success in at least one trial = $P(X \ge 1) = 1 – P(X = 0) = 1 – q^n$.
• Probability of at least $x$ successes = $P(X \ge x) = \sum\limits_{x} \binom{n}{x}p^xq^{n-x}\quad (x = x, x + 1,\cdots, n$)
• Probability of at most $x$ successes = $P(X \le x) =\sum\limits_{x} \binom{n}{x}p^x q^{n-x}\quad (x=0,1,\cdots,x)$
• If in $n$ trials, the experiment is repeated $N$ times, the expected frequencies are $N\cdot P(x)$ for $x = 0, 1, 2, 3, \cdots, n$.

Solved Binomial Distribution Questions

Question 1: A die is rolled 5 times and a 5 or 6 is considered a success. Find the probability of (i) no success, (ii) at least 2 successes, (iii) at least one but not more than 3 successes.

Solution:

The Sample Space is $S=\{1, 2, 3, 4, 5, 6\}$. Since the occurrence of 5 or 6 is considered a success, therefore, $p=\frac{2}{6}=\frac{1}{3} \Rightarrow q=1-p = 1-\frac{1}{3} = \frac{2}{3}$.

(i) No success

$n=5, p=\frac{1}{3}, q=\frac{2}{3}$, x=0$

\begin{align*}
P(X=x) &= \binom{n}{x}p^x q^{n-x}\\
P(X=0) &= \binom{5}{0} \left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^5=0.1316
\end{align*}

(ii) At least 2 successes

\begin{align*}
P(X \ge 2) & = 1 – P(X<2)\\
&= 1 – [P(X=0) + P(X=1)]\\
&= 1- [0.13168 + 0.3292] = 0.5391
\end{align*}

(iii) At least one but not more than 3 successes

\begin{align*}
P(1 \le x \le 3) &= P(X=1) + P(X=2) + P(X=3)\\
&= 0.3292 + 0.3292 + 0.1646 = 0.823
\end{align*}

Question 2: Find the probability of getting (i) exactly 4 heads and (ii) not more than 4 heads when 6 coins are tossed.

Solution:

From the given information, $n = 6, x = 4, p = q = \frac{1}{2}$

(i) Exactly 4 heads

\begin{align*}
P(X=x) &= \binom{n}{x} p^x q^{n-x}\\
&= \binom{6}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^{6-4} = 0.234
\end{align*}

(ii) Not more than 4 heads

\begin{align*}
P(X\le 4) & = 1 – p(X\ge 4)\\
&= 1 – P(X=4) + P(X=5) + P(X=6)
\end{align*}

Question 3: If 60% of the voters in a large district prefer candidate-A, what is the probability that in a sample of 12 voters, exactly 7 will prefer A?

Solution:

From given information in the questions, $p=06, q=0.4, n=12, x=7$

\begin{align*}
P(X=x)&= \binom{n}{x}p^x q^{n-x}\\
P(X=7) &= \binom{12}{7} (0.6)^7(0.4)^5&= 0.227
\end{align*}

Question 4: The probability that a patient recovers from a delicate heart operation is 0.9. What is the probability that exactly 5 of the next 7 patients having this operation survive?

Solution:

From the given information in the question, $n=7, x=5, p=0.9, q=0.10$

\begin{align*}
P(X=x)&= \binom{n}{x}p^x q^{n-x}\\
P(X=5) &= \binom{7}{5}(0.9)^5(0.1)^2 = 0.124
\end{align*}

Question 5: The incidence of occupational disease in an industry is such that the workmen have a 20% chance of suffering from it. What is the probability that out of 6 workmen (i) not more than 2, and (ii) 4 or more will catch the disease?

Solution:

From the given information in the questions

Probability of suffering from occupational disease = $\frac{20}{100}=\frac{1}{5}=0.20$

Probability of not suffering from occupational disease = $1 – \frac{1}{5} = \frac{4}{5}=0.80$

(i) Probability that out of 6 workers, not more than two will suffer

\begin{align*}
P(X\le 2) &= \binom{6}{0}\left(\frac{4}{5}\right)^0 \left(\frac{1}{5}\right)^6 + \binom{6}{1}\left(\frac{4}{5}\right)^1 \left(\frac{1}{5}\right)^5 + \binom{6}{2}\left(\frac{4}{5}\right)^2 \left(\frac{1}{5}\right)^4\\
&=0.01696
\end{align*}

(ii) Probability that out of 6 workers, 4 or more will suffer

\begin{align*}
P(X\ge 4) &= \binom{6}{4}\left(\frac{4}{5}\right)^4 \left(\frac{1}{5}\right)^2 + \binom{6}{5}\left(\frac{4}{5}\right)^5 \left(\frac{1}{5}\right)^1 + \binom{6}{6}\left(\frac{4}{5}\right)^6 \left(\frac{1}{5}\right)^0\\
&=0.90112
\end{align*}

Question 6: A multiple-choice has 15 questions, each with 4 possible answers of which only 1 is the correct answer. What is the probability that sheer guesswork yields from 5 to 10 correct answers?

Solution:

Probability of answering any question correctly: $p=\frac{1}{4}=0.25$

Probability of answering any question wrongly: $q=\frac{3}{4}=0.75$

\begin{align*}
P(5 \le x \le 10) &= P(X=5) + P(X=6) + \cdots + P(X=10)\\
&=\binom{15}{5}\left(\frac{1}{4}\right)^5\left(\frac{3}{4}\right)^{10}+\cdots + \binom{15}{10}\left(\frac{1}{4}\right)^5\left(\frac{3}{4}\right)^{5} \\
&= 0.31339
\end{align*}

Question 7: A commuter drivers to work each morning. The route she takes each day includes ten stoplights. Assume the probability each stoplight is red when she gets to it is 0.2 and that these stoplights (trials) are independent. What is the distribution of $X$, the number of times she must stop for a red light on her way to work? Evaluates $P(X=0) and $P(X<3).

Solution:

The distribution of $X$ is binomial because trials are independent. The probability of getting red spotlight (success) is 0.2 which remains the same, the number of trials is fixed ($n=10$).

The further information given in the Question is: $n=10, p=0.2, q=0.8$

\begin{align*}
P(X=0)&=\binom{10}{0} (0.2)^0(0.8)^{10} = 0.10737\\
P(X<3) &=\binom{10}{0} (0.2)^0(0.8)^{10} + \binom{10}{1} (0.2)^1(0.8)^{9} + \binom{10}{2} (0.2)^2(0.8)^{8} = 0.6777
\end{align*}

Application of Binomial Probability Distribution

• Quality Control:
  • Assessing Product Reliability: Manufacturers use binomial distribution to estimate the probability of defective products in a batch, helping them maintain quality standards.
  • Predicting Failure Rates: By analyzing past data, companies predict the likelihood of equipment failure using a binomial probability distribution, aiding in preventive maintenance and reducing downtime.

• Genetics:
  • Predicting Inheritance Patterns: In genetics, Binomial distribution helps to predict the probability of offspring inheriting specific traits based on parental genotypes.
  • Analyzing Genetic Mutations: Binomial distribution is used to study the frequency of genetic mutations in populations.

• Medicine:
  • Clinical Trials: Binomial distribution is essential for designing and analyzing clinical trials, assessing the effectiveness of treatments, and determining the probability of side effects.
  • Epidemiology: Binomial distribution helps to model the spread of infectious diseases and predict outbreak risks.

• Finance:
  • Risk Assessment: Financial institutions use Binomial Probability Distribution to assess the risk of loan defaults or investment failures.
  • Option Pricing: Binomial probability distribution is a key component of option pricing models, helping to determine the fair value of options contracts.

• Social Sciences:
  • Survey Analysis: Binomial distribution is used to analyze survey data, such as predicting voter behavior or public opinion on specific issues.
  • Market Research: Binomial Probability Distribution helps businesses to understand consumer preferences and predict market trends.

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