Binomial Distribution (Class 12): VSA and Short Answer Questions with Solutions
Step-by-step solutions to 14 Very Short Answer (2 marks) and 11 Short Answer (4 marks) questions on the binomial distribution: probability distribution, mean, variance, S.D., finding \(n\) and \(p\), and probabilities of coin, card, bolt and bomb problems. Each solution uses \(P(X=r)=\binom nr p^rq^{n-r}\) and has a boxed final answer. Useful for CBSE Class 12 board exam, JEE and other competitive exam preparation.
Consider \(n\) independent Bernoulli trials, each with probability of success \(p\) and probability of failure \(q=1-p\). Let \(X\) be the number of successes. Then \(X\) takes the values \(0,1,2,\dots,n\), and by the binomial law
\[P(X=r)=\binom{n}{r}p^{r}q^{\,n-r},\qquad r=0,1,2,\dots,n.\]| \(X\) | \(0\) | \(1\) | \(2\) | \(\cdots\) | \(n\) |
|---|---|---|---|---|---|
| \(P(X)\) | \(\binom n0q^n\) | \(\binom n1pq^{n-1}\) | \(\binom n2p^2q^{n-2}\) | \(\cdots\) | \(\binom nnp^n\) |
The probabilities are the terms of the expansion of \((q+p)^n=1\). This is written \(X\sim B(n,p)\).
For a binomial variable, \(\mathrm{Var}(x)=npq=np(1-p)=n(p-p^2)\). Let \(f(p)=n(p-p^2)\), \(0\le p\le1\).
\[f'(p)=n(1-2p)=0\ \Rightarrow\ p=\frac12,\qquad f''(p)=-2n<0.\]So \(f\) has a maximum at \(p=\frac12\), and the maximum value is \(n\cdot\frac12\cdot\frac12=\frac n4\).
A set of numbers is a probability distribution if (a) every value is between 0 and 1, and (b) the values add up to 1.
(i)All four values lie between 0 and 1, and
\[0.26+0.26+0.26+0.22=1.00.\]\(f(1)=\frac24,\ f(2)=\frac34,\ f(3)=\frac44=1,\ f(4)=\frac54>1\). Also \(f(1)+f(2)+f(3)+f(4)=\dfrac{2+3+4+5}{4}=\dfrac{14}{4}\ne1\).
For a binomial distribution, mean \(=np\), variance \(=npq\), and \(0 Variance \(=(\text{S.D.})^2=16\). Then \(np=16\) and \(npq=16\), which gives \(q=1\), i.e. \(p=0\). But then \(np=0\ne16\). Equivalently, the variance cannot be equal to the mean. \(np=4\) and \(npq=3\). Dividing, \(q=\frac34\), so \(p=\frac14\) and \(n=\dfrac{4}{1/4}=16\). These are valid values (\(n\) is a positive integer and \(0
Let \(X\) be the number of tails. Then \(X\sim B\left(7,\frac12\right)\) and \(P(X=r)=\binom7r\left(\frac12\right)^7=\dfrac{\binom7r}{128}\). The values are \(\binom70=1,\ \binom71=7,\ \binom72=21\).
(i) Exactly two tails\[P(X=2)=\frac{21}{128}.\](ii) At most two tails\[P(X\le2)=\frac{1+7+21}{128}=\frac{29}{128}.\](iii) At least two tails\[P(X\ge2)=1-P(X\le1)=1-\frac{1+7}{128}=\frac{120}{128}=\frac{15}{16}.\]Let \(X\) be the number of heads. Then \(X\sim B\left(8,\frac12\right)\) and \(P(X=r)=\dfrac{\binom8r}{256}\).
\[P(X\ge6)=\frac{\binom86+\binom87+\binom88}{256}=\frac{28+8+1}{256}=\frac{37}{256}.\](ii) Six coins are tossed simultaneously. Find the probability of getting (a) 3 heads, (b) no head, (c) at least one head, (d) not more than 3 heads.
\(X\sim B\left(8,\frac12\right)\), \(P(X=r)=\dfrac{\binom8r}{256}\).
\[P(X\ge3)=1-P(X\le2)=1-\frac{1+8+28}{256}=1-\frac{37}{256}=\frac{219}{256}.\]\(X\sim B\left(6,\frac12\right)\), \(P(X=r)=\dfrac{\binom6r}{64}\), with \(\binom60=1,\ \binom61=6,\ \binom62=15,\ \binom63=20\).
- (a) \(P(X=3)=\dfrac{20}{64}=\dfrac5{16}\)
- (b) \(P(X=0)=\dfrac1{64}\)
- (c) \(P(X\ge1)=1-\dfrac1{64}=\dfrac{63}{64}\)
- (d) \(P(X\le3)=\dfrac{1+6+15+20}{64}=\dfrac{42}{64}=\dfrac{21}{32}\)
Let \(X\) be the number of boys. Each birth is independent with \(p=\frac12\), so \(X\sim B\left(5,\frac12\right)\) and \(P(X=r)=\dfrac{\binom5r}{32}\).
(i) At least one boy\[P(X\ge1)=1-P(X=0)=1-\frac1{32}=\frac{31}{32}.\](ii) At least one boy and one girlThis is the complement of "all boys" or "all girls":
\[P=1-P(X=5)-P(X=0)=1-\frac1{32}-\frac1{32}=\frac{30}{32}=\frac{15}{16}.\]Mean \(np=4\) and variance \(npq=3\). Dividing the second equation by the first,
\[q=\frac34,\qquad p=1-q=\frac14,\qquad n=\frac{4}{p}=16.\]Each answer is guessed correctly with probability \(p=\frac12\). Let \(X\) be the number of correct answers, so \(X\sim B\left(8,\frac12\right)\) and \(P(X=r)=\dfrac{\binom8r}{256}\).
\[P(X\ge5)=\frac{\binom85+\binom86+\binom87+\binom88}{256}=\frac{56+28+8+1}{256}=\frac{93}{256}.\]Let \(X\) be the number of heads. Then \(X\sim B\left(10,\frac12\right)\) and \(P(X=r)=\dfrac{\binom{10}{r}}{1024}\). The needed values are \(\binom{10}6=210,\ \binom{10}7=120,\ \binom{10}8=45,\ \binom{10}9=10,\ \binom{10}{10}=1\).
(i) Exactly six heads\[P(X=6)=\frac{210}{1024}=\frac{105}{512}.\](ii) At least six heads\[P(X\ge6)=\frac{210+120+45+10+1}{1024}=\frac{386}{1024}=\frac{193}{512}.\](iii) At most six heads\[P(X\le6)=1-P(X\ge7)=1-\frac{120+45+10+1}{1024}=1-\frac{176}{1024}=\frac{848}{1024}=\frac{53}{64}.\]Because the cards are replaced, the draws are independent. In each draw, \(p=P(\text{spade})=\dfrac{13}{52}=\dfrac14\) and \(q=\dfrac34\). Let \(X\) be the number of spades in 5 draws, so \(X\sim B\left(5,\frac14\right)\).
(i) All five are spades\[P(X=5)=\binom55\left(\frac14\right)^5=\frac{1}{1024}.\](ii) Only 3 spades\[P(X=3)=\binom53\left(\frac14\right)^3\left(\frac34\right)^2=10\times\frac{1}{64}\times\frac{9}{16}=\frac{90}{1024}=\frac{45}{512}.\](iii) No spade\[P(X=0)=\left(\frac34\right)^5=\frac{243}{1024}.\]- The experiment consists of a fixed number \(n\) of trials.
- Each trial has exactly two outcomes, "success" and "failure".
- The trials are independent.
- The probability of success \(p\) is the same in every trial (and \(q=1-p\)).
If \(X\) denotes the number of successes in the \(n\) trials, then \(X\) can be \(0\) (no success), \(1, 2, \dots\) up to \(n\) (all trials succeed). So the values of \(X\) are \(0,1,2,\dots,n\), with \(P(X=r)=\binom nr p^rq^{n-r}\).
ExampleA fair coin is tossed 3 times, and \(X\) is the number of heads. Here \(n=3,\ p=\frac12\), so the values of \(X\) are \(0,1,2,3\).
| \(X\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| \(P(X)\) | \(\frac18\) | \(\frac38\) | \(\frac38\) | \(\frac18\) |
Let \(X\) be a discrete random variable taking the values \(x_1,x_2,\dots,x_k\). Its probability distribution is the list of these values together with their probabilities \(p_i=P(X=x_i)\), where
\[p_i\ge0\quad\text{for every } i,\qquad \sum_{i=1}^{k}p_i=1.\]Binomial caseFor \(X\sim B(n,p)\) with \(q=1-p\):
| \(X\) | \(0\) | \(1\) | \(2\) | \(\cdots\) | \(r\) | \(\cdots\) | \(n\) |
|---|---|---|---|---|---|---|---|
| \(P(X)\) | \(q^n\) | \(\binom n1pq^{n-1}\) | \(\binom n2p^2q^{n-2}\) | \(\cdots\) | \(\binom nr p^rq^{n-r}\) | \(\cdots\) | \(p^n\) |
The probabilities add up to \(\sum_{r=0}^n\binom nr p^rq^{n-r}=(q+p)^n=1\), as required.
Let \(X\sim B(n,p)\), \(q=1-p\), so \(P(X=r)=\binom nr p^rq^{n-r}\).
MeanUsing \(r\binom nr=n\binom{n-1}{r-1}\),
\[E(X)=\sum_{r=1}^{n}r\binom nr p^rq^{n-r}=np\sum_{r=1}^{n}\binom{n-1}{r-1}p^{r-1}q^{(n-1)-(r-1)}=np(q+p)^{n-1}=np.\]VarianceUsing \(r(r-1)\binom nr=n(n-1)\binom{n-2}{r-2}\),
\[E[X(X-1)]=\sum_{r=2}^{n}r(r-1)\binom nr p^rq^{n-r}=n(n-1)p^2(q+p)^{n-2}=n(n-1)p^2.\]\[\begin{aligned}\mathrm{Var}(X)&=E[X(X-1)]+E(X)-[E(X)]^2\\&=n(n-1)p^2+np-n^2p^2=np-np^2=np(1-p)=npq.\end{aligned}\]Standard deviation\[\text{S.D.}=\sqrt{\mathrm{Var}(X)}=\sqrt{npq}.\]A binomial variable \(X\sim B(n,p)\) is the sum \(X=X_1+X_2+\cdots+X_n\) of \(n\) independent Bernoulli variables, where each \(X_i\) takes the value 1 with probability \(p\) and 0 with probability \(q=1-p\). For each \(X_i\),
\[E(X_i)=p,\qquad E(X_i^2)=p,\qquad \mathrm{Var}(X_i)=p-p^2=pq.\]The variances of independent variables add, so
\[\mathrm{Var}(X)=npq.\]Variance cannot exceed n/4Since \(q=1-p\), consider
\[\frac14-pq=\frac14-p(1-p)=p^2-p+\frac14=\left(p-\frac12\right)^2\ge0.\]Hence \(pq\le\dfrac14\), and multiplying by \(n>0\), \(\ npq\le\dfrac n4\). Equality holds only for \(p=\frac12\).
For \(X\sim B(n,p)\) with \(q=1-p\) and \(0 Then Since \(0 Let \(X\sim B(n,p)\), \(q=1-p\). Its mean is \(np\) and its variance is \(npq\). Now \(0\le q\le1\) because \(q\) is a probability. Hence because \(np\ge0\). The inequality is strict (variance < mean) whenever \(0 \(p=P(\text{defective})=\frac{20}{100}=\frac15\), \(q=\frac45\), \(n=4\). Let \(X\) be the number of defective bolts, \(P(X=r)=\binom4r\left(\frac15\right)^r\left(\frac45\right)^{4-r}\). \(np=4\) and \(npq=2\), so \(q=\frac12\), \(p=\frac12\) and \(n=8\). Then \(P(X=r)=\dfrac{\binom8r}{256}\). \(np=4\) and \(npq=\frac43\), so \(q=\frac13\), \(p=\frac23\) and \(n=6\). \(np=12\) and \(npq=3\), so \(q=\frac14\), \(p=\frac34\) and \(n=16\). \(p=0.4\), \(q=0.6\), \(n=3\). Let \(X\) be the number of students who graduate, \(P(X=r)=\binom3r(0.4)^r(0.6)^{3-r}\). \(p=0.2=\frac15\), \(q=0.8=\frac45\), \(n=6\). Let \(X\) be the number of direct hits. The bridge is destroyed if \(X\ge2\). Let \(p\) be the probability of success and \(q=1-p\). For 7 trials, \(P(X=4)=P(X=3)\): Since \(\binom74=\binom73=35\), dividing by \(35p^3q^3\) gives \(p=q\). So \(p=q=\frac12\). For 5 trials with \(p=\frac12\): Mean \(np=3\) and variance \(npq=(\sqrt2)^2=2\). Dividing, \(q=\frac23\), so \(p=\frac13\) and \(n=9\). \(np=4\) and \(npq=2\), so \(q=\frac12\), \(p=\frac12\) and \(n=8\). The variate assumes a non-zero value when \(X\ne0\):\(r\) \(P(X=r)\) \(0\) \(\left(\frac45\right)^4=\frac{256}{625}\) \(1\) \(4\cdot\frac15\cdot\left(\frac45\right)^3=\frac{256}{625}\) \(2\) \(6\cdot\frac1{25}\cdot\frac{16}{25}=\frac{96}{625}\)
(ii) The mean and variance of a binomial distribution are 4 and \(\frac43\) respectively. Find \(P(X\ge1)\).
(iii) Find the binomial distribution for which the mean and variance are 12 and 3 respectively.
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