Binomial Distribution
Binomial Distribution: Definition, Moments, MGF and Solved Examples
A Binomial Distribution is a discrete probability distribution that gives the probability of obtaining exactly \(x\) successes in \(n\) independent Bernoulli trials, where the probability of success remains constant.
If
\(X \sim \operatorname{Bin}(n,p)\),
then the probability mass function is
\[ P(X=x)=\binom{n}{x}p^xq^{\,n-x}, \qquad x=0,1,2,\ldots,n, \]
where
- \(n\) = number of trials,
- \(p\) = probability of success,
- \(q=1-p\) = probability of failure,
- \(X\) = number of successes.
Conditions for Binomial Distribution
- The number of trials \(n\) is fixed.
- Each trial has only two possible outcomes: success or failure.
- The probability of success \(p\) remains constant.
- The trials are independent.
Moments of Binomial Distribution
Let
\[ X\sim \operatorname{Bin}(n,p), \qquad q=1-p. \]
First Moment: Mean
\[ \boxed{E(X)=np} \]
Second Raw Moment
Since
\[ E[X(X-1)]=n(n-1)p^2, \]
and
\[ X^2=X(X-1)+X, \]
we obtain
\[ \boxed{E(X^2)=n(n-1)p^2+np}. \]
Variance
\[ \boxed{\operatorname{Var}(X)=npq} \]
Therefore, the standard deviation is
\[ \boxed{\sigma=\sqrt{npq}}. \]
Third Central Moment
\[ \boxed{\mu_3=npq(q-p)} \]
or
\[ \boxed{\mu_3=npq(1-2p)}. \]
Coefficient of Skewness
\[ \boxed{ \gamma_1= \frac{q-p}{\sqrt{npq}} } \]
Fourth Central Moment
\[ \boxed{ \mu_4=3(npq)^2+npq(1-6pq) } \]
Kurtosis
\[ \boxed{ \beta_2= 3+\frac{1-6pq}{npq} } \]
The excess kurtosis is
\[ \boxed{ \gamma_2= \frac{1-6pq}{npq} } \]
Moment Generating Function
The moment generating function of a random variable \(X\) is defined as
\[ M_X(t)=E(e^{tX}). \]
For a binomial random variable,
\[ M_X(t) = \sum_{x=0}^{n} e^{tx} \binom{n}{x} p^xq^{n-x}. \]
Since
\[ e^{tx}p^x=(pe^t)^x, \]
we have
\[ M_X(t) = \sum_{x=0}^{n} \binom{n}{x} (pe^t)^xq^{n-x}. \]
Using the binomial theorem,
\[ \boxed{ M_X(t)=(q+pe^t)^n } \]
Solved Examples
Example 1: Exact Probability
A fair coin is tossed 8 times. Find the probability of obtaining exactly 5 heads.
Show Solution
Let \(X\) denote the number of heads.
\[ n=8,\qquad p=\frac12,\qquad q=\frac12. \]
We require \(P(X=5)\).
\[ P(X=5) = \binom85 \left(\frac12\right)^5 \left(\frac12\right)^3. \]
Since \[ \binom85=56, \]
\[ P(X=5) = 56\left(\frac12\right)^8 = \frac{56}{256}. \]
Therefore, \[ \boxed{P(X=5)=0.21875}. \]
Example 2: At Least One Success
A machine produces defective items with probability \(0.05\). If 10 items are selected independently, find the probability that at least one is defective.
Show Solution
Let \[ X\sim \operatorname{Bin}(10,0.05). \]
Thus \[ p=0.05,\qquad q=0.95. \]
We need \[ P(X\ge 1). \]
Using the complement rule,
\[ P(X\ge1)=1-P(X=0). \]
\[ P(X=0) = (0.95)^{10}. \]
Therefore, \[ P(X\ge1) = 1-(0.95)^{10}. \]
Since \[ (0.95)^{10}\approx0.59874, \]
\[ \boxed{P(X\ge1)\approx0.40126}. \]
Example 3: Mean and Variance
Suppose \[ X\sim\operatorname{Bin}(20,0.3). \] Find the mean, variance and standard deviation.
Show Solution
Here \[ n=20,\qquad p=0.3,\qquad q=0.7. \]
Mean:
\[ E(X)=np = 20(0.3) = \boxed{6}. \]
Variance:
\[ \operatorname{Var}(X) = npq = 20(0.3)(0.7) = \boxed{4.2}. \]
Standard deviation:
\[ \sigma=\sqrt{4.2}\approx2.049. \]
Therefore, \[ \boxed{\sigma\approx2.049}. \]
Example 4: Finding \(n\) and \(p\)
A binomial random variable has mean \(6\) and variance \(2.4\). Find \(n\) and \(p\).
Show Solution
For a binomial distribution,
\[ np=6 \]
and
\[ npq=2.4. \]
Dividing the second equation by the first,
\[ q=\frac{2.4}{6}=0.4. \]
Therefore,
\[ p=1-q=0.6. \]
Since
\[ np=6, \]
\[ n(0.6)=6. \]
Hence,
\[ n=10. \]
Therefore, \[ \boxed{n=10,\qquad p=0.6}. \]
Example 5: At Least Four Successes
The probability that a student solves a problem correctly is \(0.7\). If the student attempts 6 independent problems, find the probability that at least 4 are solved correctly.
Show Solution
Let \[ X\sim\operatorname{Bin}(6,0.7). \]
Then \[ q=0.3. \]
We require \[ P(X\ge4) = P(X=4)+P(X=5)+P(X=6). \]
For \(X=4\),
\[ P(X=4) = \binom64(0.7)^4(0.3)^2 = 15(0.2401)(0.09) = 0.324135. \]
For \(X=5\),
\[ P(X=5) = \binom65(0.7)^5(0.3) = 0.302526. \]
For \(X=6\),
\[ P(X=6) = (0.7)^6 = 0.117649. \]
Therefore,
\[ P(X\ge4) = 0.324135+0.302526+0.117649. \]
Hence, \[ \boxed{P(X\ge4)=0.74431}. \]
Example 6: Identifying a Distribution from its MGF
Suppose the moment generating function of \(X\) is
\[ M_X(t)=(0.4+0.6e^t)^5. \]
Identify the distribution and find its mean and variance.
Show Solution
The MGF of a binomial random variable is
\[ M_X(t)=(q+pe^t)^n. \]
Comparing with
\[ (0.4+0.6e^t)^5, \]
we obtain
\[ n=5,\qquad p=0.6,\qquad q=0.4. \]
Thus \[ X\sim\operatorname{Bin}(5,0.6). \]
The mean is
\[ E(X)=np=5(0.6)=\boxed{3}. \]
The variance is
\[ \operatorname{Var}(X) = npq = 5(0.6)(0.4). \]
Therefore, \[ \boxed{\operatorname{Var}(X)=1.2}. \]
Example 7: Third Moment and Skewness
Let \[ X\sim\operatorname{Bin}(50,0.2). \] Find the third central moment and coefficient of skewness.
Show Solution
Here \[ n=50,\qquad p=0.2,\qquad q=0.8. \]
The third central moment is
\[ \mu_3=npq(q-p). \]
Therefore,
\[ \mu_3 = 50(0.2)(0.8)(0.8-0.2). \]
\[ \mu_3 = 50(0.16)(0.6) = \boxed{4.8}. \]
The coefficient of skewness is
\[ \gamma_1 = \frac{q-p}{\sqrt{npq}}. \]
Now,
\[ npq = 50(0.2)(0.8) = 8. \]
Hence,
\[ \gamma_1 = \frac{0.6}{\sqrt8} \approx0.2121. \]
Therefore, \[ \boxed{\gamma_1\approx0.2121}. \]
Since the skewness is positive, the distribution is slightly right-skewed.
Example 8: Communication System
A communication system transmits each bit correctly with probability \(0.98\). A block contains 20 independently transmitted bits. Find:
(i) \(P(X=20)\)
(ii) \(P(X\ge19)\)
where \(X\) denotes the number of correctly transmitted bits.
Show Solution
Here \[ X\sim\operatorname{Bin}(20,0.98), \]
so
\[ p=0.98,\qquad q=0.02. \]
(i) Probability that all 20 bits are correct
\[ P(X=20) = \binom{20}{20}(0.98)^{20}. \]
Therefore, \[ P(X=20) = (0.98)^{20} \approx0.6676. \]
Hence, \[ \boxed{P(X=20)\approx0.6676}. \]
(ii) Probability that at least 19 bits are correct
\[ P(X\ge19) = P(X=19)+P(X=20). \]
Now, \[ P(X=19) = \binom{20}{19} (0.98)^{19}(0.02). \]
Thus, \[ P(X=19) = 20(0.98)^{19}(0.02) \approx0.2725. \]
Therefore,
\[ P(X\ge19) = 0.2725+0.6676. \]
Hence, \[ \boxed{P(X\ge19)\approx0.9401}. \]
Important Formula Summary
| Quantity | Formula |
|---|---|
| Probability Mass Function | \(\displaystyle P(X=x)=\binom nxp^xq^{n-x}\) |
| Mean | \(\displaystyle np\) |
| Variance | \(\displaystyle npq\) |
| Standard Deviation | \(\displaystyle \sqrt{npq}\) |
| MGF | \(\displaystyle (q+pe^t)^n\) |
| Second Raw Moment | \(\displaystyle n(n-1)p^2+np\) |
| Third Central Moment | \(\displaystyle npq(q-p)\) |
| Skewness | \(\displaystyle \frac{q-p}{\sqrt{npq}}\) |
| Fourth Central Moment | \(\displaystyle 3(npq)^2+npq(1-6pq)\) |
| Excess Kurtosis | \(\displaystyle \frac{1-6pq}{npq}\) |
Important: When \(p=q=\frac12\), the binomial distribution is symmetric and its coefficient of skewness is zero.
Example 9: Finding an Unknown Probability
A binomial random variable \(X\) has parameters \(n=10\) and \(p\). If
\[ P(X=0)=\frac{1}{1024}, \]
find \(p\), the mean, and the variance.
Show Solution
For a binomial random variable,
\[ P(X=x)=\binom{n}{x}p^xq^{n-x}, \qquad q=1-p. \]
For \(X=0\),
\[ P(X=0)=q^{10}. \]
Given
\[ q^{10}=\frac{1}{1024}. \]
Since
\[ 1024=2^{10}, \]
we get
\[ q^{10}=\left(\frac12\right)^{10}. \]
Therefore,
\[ q=\frac12. \]
Hence,
\[ p=1-q=\frac12. \]
Mean:
\[ E(X)=np=10\left(\frac12\right)=5. \]
Variance:
\[ \operatorname{Var}(X)=npq = 10\left(\frac12\right)\left(\frac12\right) = 2.5. \]
Therefore,
\[ \boxed{ p=\frac12,\qquad E(X)=5,\qquad \operatorname{Var}(X)=2.5 } \]
Example 10: Probability Between Two Values
A multiple-choice examination contains 12 questions. Each question has four alternatives, only one of which is correct. A student answers every question by random guessing. Find the probability that the student answers at least 3 but not more than 5 questions correctly.
Show Solution
Let \(X\) denote the number of correct answers.
Since the probability of answering one question correctly is
\[ p=\frac14, \]
we have
\[ q=1-p=\frac34. \]
Therefore,
\[ X\sim\operatorname{Bin}\left(12,\frac14\right). \]
We need
\[ P(3\leq X\leq5). \]
Thus,
\[ P(3\leq X\leq5) = P(X=3)+P(X=4)+P(X=5). \]
Using the binomial formula,
\[ P(3\leq X\leq5) = \sum_{x=3}^{5} \binom{12}{x} \left(\frac14\right)^x \left(\frac34\right)^{12-x}. \]
Hence,
\[ \boxed{ P(3\leq X\leq5) = \sum_{x=3}^{5} \binom{12}{x} \left(\frac14\right)^x \left(\frac34\right)^{12-x} } \]
Example 11: Finding Parameters from Mean and Standard Deviation
For a binomial random variable \(X\), the mean is \(8\) and the standard deviation is \(2\). Determine \(n\) and \(p\). Also find
\[ P(X=10). \]
Show Solution
For a binomial distribution,
\[ E(X)=np. \]
Given that the mean is \(8\),
\[ np=8. \]
The standard deviation is \(2\), so the variance is
\[ \sigma^2=4. \]
For the binomial distribution,
\[ npq=4. \]
Dividing the variance equation by the mean equation,
\[ q=\frac{4}{8}=\frac12. \]
Therefore,
\[ p=1-q=\frac12. \]
Using
\[ np=8, \]
we get
\[ n\left(\frac12\right)=8. \]
Hence,
\[ n=16. \]
Now,
\[ P(X=10) = \binom{16}{10} \left(\frac12\right)^{10} \left(\frac12\right)^6. \]
Therefore,
\[ P(X=10) = \binom{16}{10} \left(\frac12\right)^{16}. \]
Since
\[ \binom{16}{10}=8008, \]
we obtain
\[ P(X=10) = \frac{8008}{65536} \approx0.1222. \]
Thus,
\[ \boxed{ n=16,\qquad p=\frac12,\qquad P(X=10)\approx0.1222 } \]
Example 12: MGF and Higher Moments
A random variable \(X\) has moment generating function
\[ M_X(t)=\left(\frac25+\frac35e^t\right)^8. \]
Identify the distribution of \(X\) and determine
\[ E(X),\qquad E(X^2),\qquad \operatorname{Var}(X),\qquad \mu_3. \]
Show Solution
The MGF of a binomial random variable is
\[ M_X(t)=(q+pe^t)^n. \]
Comparing with
\[ M_X(t)=\left(\frac25+\frac35e^t\right)^8, \]
we get
\[ n=8,\qquad p=\frac35,\qquad q=\frac25. \]
Therefore,
\[ \boxed{ X\sim\operatorname{Bin}\left(8,\frac35\right) } \]
Mean:
\[ E(X)=np = 8\left(\frac35\right) = \frac{24}{5} = 4.8. \]
Second raw moment:
For a binomial random variable,
\[ E(X^2)=n(n-1)p^2+np. \]
Hence,
\[ E(X^2) = 8(7)\left(\frac35\right)^2 + 8\left(\frac35\right). \]
Therefore,
\[ E(X^2) = 56\left(\frac9{25}\right) + \frac{24}{5}. \]
\[ E(X^2) = 20.16+4.8 = 24.96. \]
Variance:
\[ \operatorname{Var}(X)=npq. \]
Thus,
\[ \operatorname{Var}(X) = 8\left(\frac35\right)\left(\frac25\right) = \frac{48}{25} = 1.92. \]
Third central moment:
\[ \mu_3=npq(q-p). \]
Therefore,
\[ \mu_3 = 8\left(\frac35\right)\left(\frac25\right) \left(\frac25-\frac35\right). \]
\[ \mu_3 = \frac{48}{25} \left(-\frac15\right) = -\frac{48}{125}. \]
Hence,
\[ \boxed{\mu_3=-0.384}. \]
Therefore,
\[ \boxed{ E(X)=4.8,\qquad E(X^2)=24.96,\qquad \operatorname{Var}(X)=1.92,\qquad \mu_3=-0.384 } \]
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