Differentiation (Part-38) | S N De

Binomial Distribution (Class 12): Short Answer Type Questions 12–26

Step-by-step solutions to fifteen 4-mark questions on the binomial distribution: mean and standard deviation, finding \(n\) and \(p\), expected frequencies, and probability problems on dice, coins, eggs, valves and surveys. Each solution uses \(P(X=r)=\binom nr p^rq^{n-r}\) and ends with a boxed answer. Useful for CBSE Class 12 board exam, JEE and other competitive exam preparation.

Q124 Marks

Five percent of the articles produced by a machine are defective. Find the mean and standard deviation in a lot of 100 articles.

Solution

Here \(n=100\), \(p=\frac{5}{100}=0.05\), \(q=0.95\). For a binomial distribution, mean \(=np\) and S.D. \(=\sqrt{npq}\).

\[\text{Mean}=np=100\times0.05=5.\]\[\text{S.D.}=\sqrt{npq}=\sqrt{100\times0.05\times0.95}=\sqrt{4.75}=\frac{\sqrt{19}}{2}\approx2.18.\]

Answer: Mean \(=5\), S.D. \(=\sqrt{4.75}\approx2.18\).

Q134 Marks

If the probability of a defective bolt is \(\dfrac1{10}\), find (i) mean, (ii) variance for the distribution of defective bolts in a total of 400.

Solution

Here \(n=400\), \(p=\frac1{10}\), \(q=\frac9{10}\).

(i) Mean\[np=400\times\frac1{10}=40.\](ii) Variance\[npq=400\times\frac1{10}\times\frac9{10}=36.\]

Answer: (i) Mean \(=40\), (ii) Variance \(=36\).

Q144 Marks

If the mean and variance of a certain binomial distribution are 4 and 3.2 respectively, find the probability of at least one success.

Solution

\(np=4\) and \(npq=3.2\). Dividing, \(q=\dfrac{3.2}{4}=0.8\), so \(p=0.2\) and \(n=\dfrac{4}{0.2}=20\).

\[P(X\ge1)=1-P(X=0)=1-q^n=1-(0.8)^{20}=1-\left(\frac45\right)^{20}.\]

Since \((0.8)^{20}\approx0.0115\),

Answer: \(P(X\ge1)=1-\left(\dfrac45\right)^{20}\approx0.9885\).

Q154 Marks

A binomial random variable \(X\) satisfies the relation \(9P(X=4)=P(X=2)\) when \(n=6\). Find the value of the parameter \(p\).

Solution

With \(n=6\) and \(q=1-p\), \(P(X=r)=\binom6r p^rq^{6-r}\). The given relation is

\[9\binom64p^4q^2=\binom62p^2q^4.\]

Since \(\binom64=\binom62=15\), dividing both sides by \(15p^2q^2\) (with \(0\[9p^2=q^2\ \Rightarrow\ 3p=q=1-p\ \Rightarrow\ 4p=1.\]

Answer: \(p=\dfrac14\).

Q164 Marks

Five unbiased dice are thrown together 96 times. Find the expected frequencies when throwing of 1, 2 or 3 is regarded as a success.

Solution

On one die, \(P(\text{1, 2 or 3})=\frac36=\frac12\). So \(p=q=\frac12\) and \(n=5\). The probability of \(r\) successes is \(P(r)=\binom5r\left(\frac12\right)^5=\dfrac{\binom5r}{32}\).

The expected frequency of \(r\) successes in \(N=96\) throws is \(N\cdot P(r)=96\times\dfrac{\binom5r}{32}=3\binom5r\).

No. of successes \(r\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	Total
\(\binom5r\)	1	5	10	10	5	1	32
Expected frequency	3	15	30	30	15	3	96

Answer: The expected frequencies for \(r=0,1,2,3,4,5\) are 3, 15, 30, 30, 15, 3.

Q174 Marks

Eight coins are tossed at a time for 256 times. Find the expected frequencies of the distribution of heads and tails and tabulate the result. Calculate the mean number of heads and standard deviation.

Solution

Let \(X\) be the number of heads in one toss of 8 coins. \(n=8\), \(p=q=\frac12\), so \(P(X=r)=\dfrac{\binom8r}{256}\). The expected frequency in \(N=256\) tosses is \(256\times P(X=r)=\binom8r\). If there are \(r\) heads, there are \(8-r\) tails.

Heads	0	1	2	3	4	5	6	7	8	Total
Tails	8	7	6	5	4	3	2	1	0
Expected frequency	1	8	28	56	70	56	28	8	1	256

Mean and standard deviation\[\text{Mean number of heads}=np=8\times\frac12=4.\]\[\text{S.D.}=\sqrt{npq}=\sqrt{8\times\frac12\times\frac12}=\sqrt2\approx1.414.\]

Answer: The expected frequencies are 1, 8, 28, 56, 70, 56, 28, 8, 1 for 0 to 8 heads. Mean \(=4\), S.D. \(=\sqrt2\approx1.41\).

Q184 Marks

Six unbiased dice are thrown 128 times; a throw of 1, 2 or 3 being called a success. The results are:

No. of success	0	1	2	3	4	5	6
Frequency	0	10	26	44	34	14	0

Find the expected frequencies and compare the actual mean with those of the expected distribution.

Solution

Step 1: Expected frequencies

\(P(\text{success})=\frac36=\frac12\), \(n=6\). So \(P(r)=\dfrac{\binom6r}{64}\) and the expected frequency is \(128\times\dfrac{\binom6r}{64}=2\binom6r\).

No. of successes \(r\)	0	1	2	3	4	5	6	Total
Observed frequency	0	10	26	44	34	14	0	128
Expected frequency	2	12	30	40	30	12	2	128

Step 2: Mean and S.D. of the expected distribution\[\text{Mean}=np=6\times\frac12=3,\qquad \text{S.D.}=\sqrt{npq}=\sqrt{6\times\frac14}=\sqrt{1.5}\approx1.22.\]Step 3: Mean and S.D. of the actual (observed) results

\(\sum fx=0+10+52+132+136+70+0=400\) and \(\sum fx^2=0+10+104+396+544+350+0=1404\), with \(N=128\).

\[\text{Actual mean}=\frac{400}{128}=3.125.\]\[\text{Actual variance}=\frac{1404}{128}-(3.125)^2=10.969-9.766=1.203,\quad \text{S.D.}\approx1.10.\]

	Actual	Expected
Mean	3.125	3
S.D.	1.10	1.22

Answer: The expected frequencies are 2, 12, 30, 40, 30, 12, 2. The actual mean \(3.125\) is very close to the expected mean \(3\); the actual S.D. \(1.10\) is a little smaller than the expected S.D. \(1.22\).

Q194 Marks

The overall percentage of failures in a certain examination is 40. What is the probability that out of a group of 6 candidates at least 4 passed the examination?

Solution

The probability of passing is \(p=1-0.4=0.6=\frac35\) and \(q=0.4=\frac25\). With \(n=6\), let \(X\) be the number who passed. We need \(P(X\ge4)=P(4)+P(5)+P(6)\).

\[\begin{aligned}P(4)&=\binom64(0.6)^4(0.4)^2=15\times0.1296\times0.16=0.31104,\\P(5)&=\binom65(0.6)^5(0.4)=6\times0.07776\times0.4=0.186624,\\P(6)&=(0.6)^6=0.046656.\end{aligned}\]\[P(X\ge4)=0.31104+0.186624+0.046656=0.54432.\]

Answer: \(0.54432=\dfrac{1701}{3125}\).

Q204 Marks

4 coins are tossed at a time. What is the probability that there are 2 heads and 2 tails?

Solution

Let \(X\) be the number of heads. \(n=4\), \(p=q=\frac12\). Two heads automatically means two tails.

\[P(X=2)=\binom42\left(\frac12\right)^2\left(\frac12\right)^2=\frac{6}{16}=\frac38.\]

Answer: \(\dfrac38\).

Q214 Marks

In a basket, it is known that 1 out of 5 eggs are rotten. 18 eggs are selected randomly. Find the probability that there are exactly 3 rotten eggs. Again find the probability that there are exactly 4 rotten eggs. Which one is greater?

Solution

\(p=\frac15\), \(q=\frac45\), \(n=18\). Let \(X\) be the number of rotten eggs.

\[P(X=3)=\binom{18}{3}\left(\frac15\right)^3\left(\frac45\right)^{15}=816\times\frac{4^{15}}{5^{18}}\approx0.2297,\]\[P(X=4)=\binom{18}{4}\left(\frac15\right)^4\left(\frac45\right)^{14}=3060\times\frac{4^{14}}{5^{18}}\approx0.2153.\]

To compare them exactly, take the ratio:

\[\frac{P(X=4)}{P(X=3)}=\frac{\binom{18}{4}}{\binom{18}{3}}\cdot\frac pq=\frac{15}{4}\times\frac14=\frac{15}{16}<1.\]

Answer: \(P(X=3)\approx0.2297\) and \(P(X=4)\approx0.2153\). So the probability of exactly 3 rotten eggs is greater.

Q224 Marks

5% of the electric valves are defective. 10 valves are drawn at random. Find the probability that at least two defective valves are found.

Solution

\(p=0.05\), \(q=0.95\), \(n=10\). Let \(X\) be the number of defective valves.

\[P(X\ge2)=1-P(X=0)-P(X=1)=1-(0.95)^{10}-10(0.05)(0.95)^9.\]

Factorising,

\[P(X\ge2)=1-(0.95)^9\,(0.95+0.5)=1-1.45\times(0.95)^9.\]

With \((0.95)^9\approx0.6302\), this gives \(1-1.45\times0.6302\approx1-0.9139\).

Answer: \(P(X\ge2)=1-1.45\,(0.95)^9\approx0.0861\).

Q234 Marks

The incidence of occupational disease in an industry is such that the workers have a 20% chance of suffering from it. What is the probability that out of six workers 4 or more will catch the disease?

Solution

\(p=0.2=\frac15\), \(q=0.8=\frac45\), \(n=6\). We need \(P(X\ge4)=P(4)+P(5)+P(6)\).

\[\begin{aligned}P(4)&=\binom64\left(\frac15\right)^4\left(\frac45\right)^2=\frac{15\times16}{15625}=\frac{240}{15625},\\P(5)&=\binom65\left(\frac15\right)^5\left(\frac45\right)=\frac{6\times4}{15625}=\frac{24}{15625},\\P(6)&=\left(\frac15\right)^6=\frac{1}{15625}.\end{aligned}\]\[P(X\ge4)=\frac{240+24+1}{15625}=\frac{265}{15625}=\frac{53}{3125}=0.01696.\]

Answer: \(\dfrac{53}{3125}=0.01696\).

Q244 Marks

If the probability of success in a single trial is 0.01, how many Bernoulli trials must be performed in order that the probability of at least one success is \(\dfrac12\) or more?

Solution

\(p=0.01\), \(q=0.99\). Let \(n\) trials be performed. Then

\[P(\text{at least one success})=1-P(0)=1-(0.99)^n.\]

We need \(1-(0.99)^n\ge\dfrac12\), i.e. \((0.99)^n\le\dfrac12\). Taking logarithms (the direction of the inequality reverses because \(\log0.99<0\)):

\[n\log(0.99)\le\log\frac12\ \Rightarrow\ n\ge\frac{\log2}{-\log0.99}=\frac{0.30103}{0.004365}\approx68.97.\]

Check: \((0.99)^{68}\approx0.505>0.5\), but \((0.99)^{69}\approx0.4998<0.5\).

Answer: At least 69 trials must be performed.

Q254 Marks

Suppose that half the population of a town are consumers of rice. Each of 100 investigators interviews 10 individuals to see whether they are consumers of rice. How many investigators do you expect to report that three or more but less than six people are consumers of rice?

Solution

For one investigator, \(p=\frac12\), \(n=10\). Let \(X\) be the number of rice consumers among the 10 people, so \(P(X=r)=\dfrac{\binom{10}r}{1024}\). "Three or more but less than six" means \(X=3,4\) or \(5\).

\[P(3\le X\le5)=\frac{\binom{10}3+\binom{10}4+\binom{10}5}{1024}=\frac{120+210+252}{1024}=\frac{582}{1024}\approx0.568.\]

The expected number of investigators (out of 100) is \(100\times P\):

\[100\times\frac{582}{1024}\approx56.84\approx57.\]

Answer: About 57 investigators are expected to report 3, 4 or 5 consumers.

Q264 Marks

In 10 independent throws of a defective die, the probability that an even number will appear 5 times is twice the probability that an even number will appear 4 times. Find the probability that an even number will not appear at all in 10 independent throws of the die.

Solution

Let \(p\) be the probability of an even number in one throw and \(q=1-p\). With \(n=10\), the given condition is \(P(X=5)=2P(X=4)\):

\[\binom{10}5p^5q^5=2\binom{10}4p^4q^6.\]

Since \(\binom{10}5=252\) and \(\binom{10}4=210\), dividing by \(p^4q^5\):

\[252\,p=420\,q\ \Rightarrow\ 3p=5q=5(1-p)\ \Rightarrow\ 8p=5.\]

So \(p=\frac58\) and \(q=\frac38\). An even number does not appear at all in 10 throws with probability

\[P(X=0)=q^{10}=\left(\frac38\right)^{10}=\frac{59049}{1073741824}\approx5.5\times10^{-5}.\]

Answer: \(\left(\dfrac38\right)^{10}=\dfrac{59049}{1073741824}\approx0.000055\).

Quick recap: For \(X\sim B(n,p)\): \(P(X=r)=\binom nr p^rq^{n-r}\), mean \(=np\), variance \(=npq\). Expected frequency of \(r\) successes in \(N\) experiments \(=N\cdot P(X=r)\). "At least one" is \(1-P(0)\). To find \(n\) and \(p\) from mean and variance, divide variance by mean to get \(q\).