Discrete Probability Distribution

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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The Poisson Probability Distribution is discrete and deals with events that can only take on specific, whole number values (like the number of cars passing a certain point in an hour). Poisson Probability Distribution models the probability of a given number of events occurring in a fixed interval of time or space, given a known average rate of occurrence ($\mu$). The events must be independent of each other and occur randomly.

The Poisson probability function gives the probability for the number of events that occur in a given interval (often a period of time) assuming that events occur at a constant rate during the interval.

Poisson Random Variable

The Poisson random variable satisfies the following conditions:

• The number of successes in two disjoint time intervals is independent
• The probability of success during a small time interval is proportional to the entire length of the time interval.
• The probability of two or more events occurring in a very short interval is negligible.

Apart from disjoint time intervals, the Poisson random variable is also applied to disjoint regions of space.

Applications of Poisson Probability Distribution

The following are a few of the applications of Poisson Probability Distribution:

• The number of deaths by horse kicking in the Prussian Army (it was the first application).
• Birth defects and genetic mutations.
• Rare diseases (like Leukemia, but not AIDS because it is infectious and so not independent), especially in legal cases.
• Car accidents
• Traffic flow and ideal gap distance
• Hairs found in McDonald’s hamburgers
• Spread of an endangered animal in Africa
• Failure of a machine in one month

The formula of Poisson Distribution

The probability distribution of a Poisson random variable $X$ representing the number of successes occurring in a given time interval or specified region of space is given by

\begin{align*}
P(X=x)&=\frac{e^{-\mu}\mu^x}{x!}\,\,\quad x=0,1,2,\cdots
\end{align*}

where $P(X=x)$ is the probability of $x$ events occurring, $e$ is the base of the natural logarithm (~2.71828), $\mu$ is the mean number of successes in the given time interval (or region of space), $x$ is the number of events we are interested in, and $x!$ is the factorial of $x$.

Mean and Variance of Poisson Distribution

If $\mu$ is the average number of successes occurring in a given time interval (or region) in the Poisson distribution, then the mean and the variance of the Poisson distribution are both equal to $\mu$. That is,

\begin{align*}
E(X) &= \mu\\
V(X) &= \sigma^2 =\mu
\end{align*}

A Poisson distribution has only one parameter, $\mu$ is needed to determine the probability of an event. For binomial experiments involving rare events (small $p$) and large values of $n$, the distribution of $X=$ the number of success out of $n$ trials is binomial, but it is also well approximated by the Poisson distribution with mean $\mu=np$.

When to Use Poisson Probability Distribution

The Poisson distribution is useful in various scenarios:

• Modeling Rare Events: Like accidents, natural disasters, or equipment failures.
• Counting Events in a Fixed Interval: Such as the number of customers arriving at a store in an hour, or the number of calls to a call center in a minute.
• Approximating the Binomial Distribution: When the number of trials ($n$) is large and the probability of success ($p$) is small.

It is important to note that

• The Poisson distribution is related to the exponential distribution, which models the time between events.
• It is a fundamental tool in probability theory and statistics, with applications in fields like operations research, queuing theory, and reliability engineering.

R and Data Analysis, Test Preparation MCQs

Frequently Asked Questions about Poisson Distribution

1. What is the Poisson Random Variable?
2. What is Poisson Probability Distribution?
3. Write the Formula of Poisson Probability Distribution.
4. Poisson distribution is related to what distribution?
5. Give some important applications of Poisson Distribution.
6. Describe the general situations in which Poisson distribution can be used.
7. Name the distribution that has equal mean and variance.
8. What are the required conditions for poison random variables?

A probability distribution for a discrete random variable $X$ is a list of each possible value for $X$ with the probability that $X$ will have that value when the experiment is run. The likelihood for the probability distribution of a discrete random variable is denoted by $P(X=x)$. The probability distribution of a discrete random variable is also called a discrete probability distribution.

A discrete probability distribution is a mathematical function that assigns probabilities to each possible value of a discrete random variable.

Example of Probability Distribution of a Discrete Random Variable

Let $X$ be a random variable representing the number of trials obtained when a coin is flipped three times in an experiment. The sample space of the experiment is:

$$HHH, HHT, HTH, THH, HTT, TTH, THT, TTT$$

where $T$ represents the occurrence of Tail and $H$ represents the occurrence of Head in the above experiment.

Then $X$ has 4 possible values: $0, 1, 2, 3$ for the occurrence of head or tail. The probability distribution for $X$ is given as below:

$X$	$P(X)$
0	$\frac{1}{8}$
1	$\frac{3}{8}$
2	$\frac{3}{8}$
3	$\frac{1}{8}$
Total	$1.0$

In a statistics class of 25 students are given a 5-point quiz. 3 students scored 0; 1 student scored 1, 4 students scored 2, 8 students scored 3, 6 students scored 4, and 3 students scored 5. If a student is chosen at random, and the random variable $S$ is the student’s Quiz Score then the discrete probability distribution of $S$ is

$S$	$P(S)$
0	0.12
1	0.04
2	0.16
3	0.32
4	0.24
5	0.12
Total	1.0

Note that for any discrete random variable $X$, $0\le P(X) \le 1$ and $\Sigma P(X) =1$.

Finding Probabilities from a Discrete Probability Distribution

Since a random variable can only take one value at a time, the events of a variable assuming two different values are always mutually exclusive. The probability of the variable taking on any number of different values can thus be found by simply adding the appropriate probabilities.

Mean or Expected Value of a Discrete Random Variable

The mean or expected value of a random variable $X$ is the average value that one should expect for $X$ over many trials of the experiment in the long run. The general notation of the mean or expected value of a random variable $X$ is represented as

$$\mu_x\quad \text{ or } E[X]$$

The mean of a discrete random variable is computed using the formula

$$E[X]=\mu_x = \Sigma x\cdot P(X)$$

Example 1

From the above experiment of three Coins the Expected value of the random variable $X$ is

$X$	$P(X)$	$x.P(X)$
0	$\frac{1}{8}$	$0 \times \frac{1}{8} = 0$
1	$\frac{3}{8}$	$1 \times \frac{3}{8} = \frac{3}{8}$
2	$\frac{3}{8}$	$2 \times \frac{3}{8} = \frac{6}{8}$
3	$\frac{1}{8}$	$3 \times \frac{1}{8} = \frac{3}{8}$
Total	$1.0$	$\frac{3}{2} = 1.5$

Thus if three coins are flipped a large number of times, one should expect the average number of trials (per 3 flips) to be about 1.5.

Example 2

Similarly, the mean of the random variable $S$ from the above example is

$S$	$P(S)$	$S\cdot P(S)$
0	0.12	$0 \times 0.12 = 0$
1	0.04	$1 \times 0.04 = 0.04$
2	0.16	$2 \times 0.16 = 0.32$
3	0.32	$3 \times 0.32 = 0.96$
4	0.24	$4\times 0.24 = 0.96$
5	0.12	$5 \times 0.12 = 0.60$
Total	$1.0$	$2.88$

Note that $2.88$ is the class average on the statistics quiz as well.

Variance and Standard Deviation of a Random Variable

One may be interested to find how much the values of a random variable differ from trial to trial. To measure this, one can define the variance and standard deviation for a random variable $X$. The variance of $X$ random variable is denoted by $\sigma^2_x$ while the standard deviation of the random variable $X$ is just the square root of $\sigma^2_x$. The formulas of variance and standard deviation of a random variable $X$ are:

\begin{align*}
\sigma^2_x &= \Sigma (x – \mu)^2 P(X)\\
\sigma_x &= \sqrt{\Sigma (x – \mu)^2 P(X)}
\end{align*}

Note that the standard deviation estimates the average difference between a value of $x$ and the expected value.

Calculating the Variance and Standard Deviation

The calculation of standard deviation for a random variable is similar to the calculation of weighted standard deviation in a frequency table. The $P(x)$ can be thought of as the relative frequency of $x$. The computation of variance and standard deviation of a random variable $X$ can be made using the following steps:

1. Compute $\mu_X$ (mean of the random variable)
2. Subtract the mean/average from each of the possible values of $X$. These values are called the deviations of the $X$ values.
3. Square each of the deviations calculated in the previous step.
4. Multiply each squared deviation (calculated in step 3) by the corresponding probability $P(x)$.
5. Sum the results of step 4. The variance of the random variable will be obtained representing $\sigma^2_X$.
6. Take the square root of the $\sigma^2_X$ computed in Step 5.

Importance of Discrete Probability Distributions

• Modeling Real-World Phenomena: Discrete Distributions help us understand and model random events in various fields of life such as engineering, finance, and the sciences.
• Decision Making: These distributions provide a framework for making informed decisions under uncertainty.
• Statistical Inference: These are used to make inferences about populations based on sample data.

FAQs about the Probability Distribution of a Discrete Random Variable

1. Define the probability distribution.
2. What is a random variable?
3. What is meant by an expected value or a random variable?
4. What is meant by the variance and standard deviation of a random variable?

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