Probability Distributions Exercise 7.1

Dear Class 12 Samacheer Kalvi students, here are the Text Book Solutions for Exercise 7.1 Business Maths Chapter 7 Probability Distributions for your reference and study.

You can find links to the other exercises below.

Probability Distributions Important Formulas

Solutions for Probability Distributions Exercise 7.2

Solutions for Probability Distributions Exercise 7.3

Solutions for Probability Distributions Exercise 7.4

Exercise 7.1

$\text{1. Define Binomial distribution} \\$
$\text{A random variable X is said to follow binomial distribution with parameter} \\$
$\text{n and p, if it assumes only non- negative value and its probability mass} \\$
$\text{function in given by P(X=x)} \\$
$= \left\{ \begin{array}{cl} nC_{x}p^{x}q^{n-x}, & : \ x=0,1,2...n\; q=1-p\\ 0 & : \ otherwise \end{array} \right \\$

$\text{2. Define Bernoulli trials.} \\$
$\text{A random experiment whose outcomes are of two types namely success S and } \\$
$\text{failure F, occurring with probabilities p and q respectively, is called a Bernoulli } \\$
$\text{ trial. occurring with probabilities p and q respectively, is called a Bernoulli trial.} \\$

$\text{3. Derive the mean and variance of binomial distribution.} \\$
$E(X)=\sum_{x=0}^{n}x\;nC_{x}p^{x}q^{n-x} \\$
$=\sum_{x=0}^{n}x\dfrac{n!}{x!(n-x)!}p^{x}q^{n-x} \\$
$=\sum_{x=1}^{n}x\dfrac{n (n-1)!}{x(x-1)!(n-x)!}p^{x-1} p\;q^{n-x} \\$
$=np\sum_{x=1}^{n}\dfrac{(n-1)!}{(x-1)!(n-x)!}p^{x-1} q^{n-x} \\$
$=np\sum_{x=1}^{n}(n-1)\; C_{(x-1)}\; p^{x-1} q^{n-x} \\$
$=np\; (q+p)^{n-1} \\$
$=np\; (1)^{n-1} \;Since\;p+q=1 \\$
$E(X)=np \\$
$E(X)^{2}=\sum_{x=0}^{n}x^{2}\;p(x) \\$
$=\sum_{x=0}^{n}(x^{2}-x+x)\;p(x) \\$
$=\sum_{x=0}^{n}(x^{2}-x)\; p(x)+\sum_{x=0}^{n}x \; p(x) \\$
$=\sum_{x=0}^{n}x\;(x-1)\; p(x)+np \\$
$=\sum_{x=0}^{n}x\;(x-1)\; nC_{x}p^{x}q^{n-x}+np \\$
$=\sum_{x=0}^{n}x\;(x-1)\; \dfrac{n!}{x! (n-x)!}p^{x}q^{n-x}+np \\$
$=\sum_{x=2}^{n}x\;(x-1)\; \dfrac{n (n-1)(n-2)!}{x(x-1)(x-2)!(n-x)!}p^{x-2}p^{2}q^{n-x}+np \\$

$=n (n-1)p^{2} \sum_{x=2}^{n} \dfrac{(n-2)!}{(x-2)!(n-x)!}p^{x-2}q^{n-x}+np \\$

$=n (n-1)p^{2} \sum_{x=2}^{n} (n-2)C_{(x-2)}! p^{x-2}q^{n-x}+np \\$

$=n (n-1)\;p^{2}\;(q+p)^{n-2}+np \\$
$=n (n-1)\;p^{2}\;(1)^{n-2}+np\; Since\; p+q=1 \\$
$E(X)^{2}=n (n-1)\;p^{2}+np \\$
$Mean=E(X)=np \\$
$Variance = E(X)^{2}+\left[ E(X) \right]^{2} \\$
$=n(n-1)\;p^{2}\;+np-(np)^{2} \\$
$(n^{2}-1)\;p^{2}+np-n^{2}p^{2} \\$
$n^{2}p^{2}-np^{2}+np-n^{2}p^{2} \\$
$-np^{2}+np \\$
$np(1-p) \\$
$Variance =npq \\$

$\text{4. Write down the conditions for which the binomial distribution can be used.} \\$
$\text{The Binomial distribution can be used under the following conditions:} \\$
$\text{1. The number of trials 'n' finite}\\$
$\text{2. The trials are independent of each other.} \\$
$\text{3. The probability of success 'p' is constant for each trial.} \\$
$\text{4. In every trial there are only two possible outcomes - success or failure.} \\$

$\text{5. Mention the properties of binomial distribution.} \\$
$\text{Properties of Binomial distribution:} \\$
$\text{1. Binomial distribution is symmetrical if p=q= 0.5. It is skew} \\$
$\text{symmetric if} \; p \neq q.\text{It is positively skewed if} \\$
$p < 0.5 \; \text{and it is negatively skewed if}\; p > 0.5} \\$
$\text{2. For Binomial distribution, variance is less than mean} \\$
$Variance \; npq = (np)q < np \\$

$6. If \; 5\% \; \text{of the items produced turn out to be defective,then find } \\$
$\text{ out the probability that out of 20 items selected at random there are} \\$
$\text{(i) exactly three defectives} \\$
$\text{(ii) atleast two defectives} \\$
$\text{(iii) exactly 4 defectives} \\$
$\text{(iv) find the mean and variance} \\$
$p=\frac{5}{100}=0.05 \\$
$q=1-0.05=0.95 \\$
$n=10 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3...10 \\$

$(i) \;p(x=3)=10C_{3}\;(0.05)^{3}\;(0.95)^{10-3} \\$
$=\frac{10\times 9 \times8}{3\times 2\times 1}(0.000125)(0.95)^{7} \\$
$=120(0.000125)(0.6983) \\$
$=0.0104745 \\$
$=0.0105 \\$

$(ii) p(x \ge2)=1-p(x<2 ) \\$
$=1-\bigg[ p(0)+p(1) \bigg] \\$
$1-\bigg[10C_{0}(0.05)^{0}(0.95)^{10}+10C_{1}(0.05)^{1}(0.95)^{10-1}\bigg] \\$
$1-\bigg[(1)(1)(0.5987)+10(0.05)(0.6302)\bigg] \\$
$1-\left( 0.5987+0.3151 \right) \\$
$1-0.9138 \\$
$p(x \ge2)= 0.0862 \\$

$(iii) p(x=4)=10C_{4}(0.05)^{4}(0.95)^{6} \\$
$=210(0.00000625)(0.7351) \\$
$=0.0009648 \\$
$p(x=4)=0.00096 \\$

$(iv)Mean=np\\$
$=10\times \dfrac{5}{100} \\$
$=\dfrac{1}{2} \\$
$=0.5 \\$
$Variance=npq \\$
$=10(0.05)(0.95) \\$
$=0.475 \\$

$\text{7. In a particular university} \; 40 \% \text{of the students are having news paper } \\$
$\text{ reading habit. Nine university students are selected to find their } \\$
$\text{views on reading habit. Find the probability that} \\$
$\text{(i) none of those selected have news paper reading habit} \\$
$\text{(ii) all those selected have news paper reading habit} \\$
$\text{(iii) atleast two third have news paper reading habit.} \\$

$p=40 \%=0.4 \\$
$q=1-0.4=0.6 \\$
$n=9 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3...9 \\$

$(i) p(x=0)=9C_{0}(0.4)^{0}(0.6)^{9} \\$
$=(1)(1)(0.01008) \\$
$p(x=0)=0.01008 \\$

$(ii) p(x=9)=9C_{9}(0.4)^{9}(0.6)^{0} \\$
$=(1)(0.00026)(1) \\$
$p(x=9)=0.00026 \\$

$(iii) p(x \ge \frac{2}{3} \times 9)=p(x \ge 6) \\$
$=p(x \ge 6)+p(x \ge 7)+p(x \ge 8)+p(x \ge 9) \\$
$=9C_{6}(0.4)^{6}(0.6)^{3}+9C_{7}(0.4)^{7}(0.6)^{2}\\$
$+9C_{8}(0.4)^{8}(0.6)^{1} +9C_{9}(0.4)^{9}(0.6)^{0} \\$
$=(84)(0.004096)(0.216)+36(0.0016384)(0.36)+9(0.00065536) \\$
$(0.6)+(1) (0.000262144)(1)\\$
$0.074317824+0.021233664+0.003538944+0.000262144 \\$
$p(x \ge \frac{2}{3} \times 9) =0.09935\\$

$\text{8. In a family of 3 children, what is the probability that} \\$
$\text{there will be exactly 2 girls?} \\$
$\text{Let X denote girls } \\$
$P(X)=\frac{1}{2} \\$
$q=1-\frac{1}{2}=\frac{1}{2} \\$
$n=3 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3 \\$
$p(X=2)=3C_{2}\left( \frac{1}{2} \right)^{2}\left( \frac{1}{2} \right)^{1} \\$
$=3 \times \left( \frac{1}{2} \right)^{3} \\$
$=\frac{3}{8} \\$
$p(X=2)=0.375 \\$

$\text{9. Defects in yarn manufactured by a local mill can be approximated by a} \\$
$\text{distribution with a mean of 1.2 defects for every 6 metres of length. If lengths}\\$
$\text{of 6 metres are to be inspected, find the probability of less than 2 defects.}\\$
$Mean=np=1.2\\$
$n=6 \\$
$6p=1.2 \\$
$p=0.2\\$
$q=1-0.2=0.8 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3 ...6\\$

$p(X<2)=p(0)+p(1)\\$
$=6C_{0} (0.2)^{0}(0.8)^{6}+6C_{1} (0.2)^{1}(0.8)^{5}\\$
$=(1)(1)(0.262144)+6(0.2)(0.32768)\\$
$=0.262144+0.393216\\$
$p(X<2)=0.65536\\$

$10.\; If\; 18\% \; \text{of the bolts produced by a machine are defective, }\\$
$\text{determine the probability that out of the 4 bolts chosen at random } \\$
$\text{(i) exactly one will be defective} \\$
$\text{(ii) none will be defective} \\$
$\text{(iii) atmost 2 will be defective} \\$

$p=18% =0.18\\$

$q=1-0.18=0.82 \\$
$n=4 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3,4\\$

$(i)\;p(X=1)=4C_{1} (0.18)^{1}(0.82)^{3}\\$
$=(4)(0.18)(0.551368)\\$
$=0.39688496\\$
$p(X=1)=0.3970\\$

$(ii)\;p(X=0)=4C_{0} (0.18)^{0}(0.82)^{4}\\$
$=(1)(1)(0.45212176)\\$
$=0.39688496\\$
$p(X=0)=0.4521\\$

$(iii)\;p(X\le 2)=p(0)+p(1)+p(2)\\$
$=(0.4521)(0.3970)+4C_{2}(0.18)^{2}(0.82)^{2}\\$
$=0.4521+0.3970+6(0.0324)(0.6724)\\$
$=0.4521+0.3970+0.1307145\\$
$p(X\le 2)=0.9798\$

$\text{11. If the probability of success is 0.09, how many trials are } \\$
$\text{needed to have a probability of atleast one success as}\; 1/3 \; \text{or more ?} \\$
$p=0.09\\$
$q=1-0.09=0.91 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3,...n\\$
$p(X \ge 1) \ge\frac{1}{3}\\$
$1-p(x <1) \ge\frac{1}{3} \\$
$-p(x <1) \ge \frac{1}{3}-1 \\$
$-p(x <1) \ge \frac{-2}{3} \\$
$p(x <1) \le \frac{2}{3} \\$
$p(0) \le \frac{2}{3} \\$
$nC_{0} (0.09)^{0}(0.91)^{n} \le \frac{2}{3}\\$
$(1)(1)(0.91)^{n} \le \frac{2}{3}\\$
$(0.91)^{n} \le 0.6667\; Eq\;(1)\\$
$\text{When n=1} 0.91<0.6667 \; \text{which is incorrect} \\$
$\text{When n=2} 0.8281<0.6667 \; \text{which is incorrect} \\$
$\text{When n=3} 0.7536<0.6667 \; \text{which is incorrect} \\$
$\text{When n=4} 0.6857<0.6667\; \text{which is incorrect} \\$
$\text{When n=5} 0.6240<0.6667 \; \text{which is correct} \\$
$\text{Therefore, n=5 or more trials} \\$

$\text{12. Among 28 professors of a certain department, 18 drive foreign cars } \\$
$\text{and 10 drive local made cars. If 5 of these professors are selected at } \\$
$\text{random, what is the probability that at least 3 of them drive foreign cars?} \\$
$\text{Let X be driving a foreign car.} \\$
$p=\frac{18}{28} =0.64\\$
$q=1-0.64=0.36 \\$
$n=5 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3,4,5\\$
$p(X \ge 3)=1-p(x<3)\\$
$=1-\bigg [ p(0)+p(1)+p(2) \bigg]\\$
$=1- \bigg [ 5C_{0} (0.64)^{0}(0.36)^{5}+5C_{1} (0.64)^{1}(0.36)^{4}+5C_{2} (0.64)^{2}(0.36)^{3} \bigg]\\$
$=1- \bigg [ (1)(1)(0.0060) +5(0.64)(0.0168)+10(0.4096)(0.0467)\bigg]\\$
$=1- \bigg[0.0060 +0.05376+0.1913 \bigg]\\$
$1-0.25106 \\$
$p(X \ge 3)=0.74894 \\$

$\text{13. Out of 750 families with 4 children each, how many families }\\$
$\text{would be expected to have }\\$
$\text{(i) atleast one boy}\\$
$\text{(ii) atmost 2 girls}\\$
$\text{(iii) and children of both sexes? }\\$
$\text{Assume equal probabilities for boys and girls.}\\$

$\text{Let X denote boys or girls } \\$
$p =\frac{1}{2} \\$
$q =1-\frac{1}{2} =\frac{1}{2} \\$
$n=4 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3,...n \\$
$(i) \text{p(at least 1 boy)}=p(x \ge 1) \\$
$p(x \ge 1)=1-p(x<1) \\$
$=1-p(0) \\$
$=1-\bigg[ 4C_{0}\left( \frac{1}{2} \right)^{0}\left( \frac{1}{2} \right)^{4} \bigg] \\$
$1-\left[ (1)(1)(\frac{1}{16}) \right] \\$
$1-\frac{1}{16} \\$
$=\frac{15}{16} \\$
$p(x \ge 1)=0.9375 \\$
$\text{Therefore, number of families with at least 1 boy}\;=0.9375 \times 750=703 \\$

$p(\text{at most 2 girls})= p(\text{at least 2 boys})=p(x \ge 2) \\$
$p(x \ge 2)=p(2)+p(3)+p(4) \\$
$=4C_{2}\left( \frac{1}{2} \right)^{2}\left( \frac{1}{2} \right)^{2}4C_{3}\left( \frac{1}{2} \right)^{3}\left( \frac{1}{2} \right)^{1}+ \\$
$4C_{4}\left( \frac{1}{2} \right)^{4}\left( \frac{1}{2} \right)^{0} \\$
$=6 \times \left( \frac{1}{2} \right)^{4}+4 \times \left( \frac{1}{2} \right)^{4}+1 \times \\$ $\left( \frac{1}{2} \right)^{4}\\$
$=11 \times \left( \frac{1}{2} \right)^{4} \\$
$=11 \times \left( \frac{1}{16} \right) \\$
$p(x \ge 2)=0.6875 \\$
$\text{Therefore, number of families with at most 2 girls }\;=0.6875 \times 750=516 \\$

$\text{p(children of both sexes)}=1-p(0)+p(4)\\$
$= 1- \bigg[ 4C_{0}\left( \frac{1}{2} \right)^{0}\left( \frac{1}{2}\right)^{4}+4C_{4} \left( \frac{1}{2} \right)^{4}\left(\frac{1}{2} \right)^{0} \bigg] \\$
$=1 - \bigg[ 1 \times \left( \frac{1}{2} \right)^{4}+1 \times \left( \frac{1} {2} \right)^{4} \bigg] \\$
$= 1- 2\times (\frac{1}{2})^{4}\\$
$1- 2\times \dfrac{1}{16} \\$
$1- \dfrac{1}{8} \\$
$\frac{7}{8} \\$
$0.8750 \\$
$\text{Therefore, number of families with children of both sexes}\;=0.8750 \times 750=656\\$

$\text{14. Forty percent of business travellers carry a laptop. In a sample of 15 } \\$
$\text{business travelers, (i) what is the probability that 3 will have a laptop?} \\$
$\text{(ii) what is the probability that 12 of the travelers will not have a laptop?} \\$
$\text{(iii) what is the probability that atleast three of the travelers have a laptop?} \\$
$p =40 \%=0.4 \\$
$q =1-.4=0.6 \\$
$n=15 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3,...n \\$
$p(X=3)=15C_{3}(0.4)^{3}(0.6)^{12} \\$
$=455(0.064)+(0.002177) \\$
$p(X=3)=0.0634 \\$
$\text{p(12 will not have laptop)}=\text{p(3 will have laptop)} \\$
$=15C_{3}(0.4)^{3}(0.6)^{12} \\$
$=455(0.064)+(0.002177) \\$
$p(X=3)=0.0634 \\$
$p(X \ge 3)=1-p(x<3) \\$
$=1-\left[ p(0)+p(1)+p(2) \right] \\$
$=1- \bigg[ 15C_{0}(0.4)^{0}(0.6)^{15}+15C_{1}(0.4)^{1}(0.6)^{14}+15C_{2}(0.4)^{2}(0.6)^{13} \bigg]\\$
$1- \bigg[ (1)(1)(0.00047)+15(0.4)(0.000784)+105(0.16)(0.00131) \bigg]\\$
$1-\left( 0.00047+0.004704+0.022008 \right)\\$
$p(X \ge 3)=0.9728\\$

$\text{15. A pair of dice is thrown 4 times. If getting a doublet is considered } \\$
$\text{a success, find the probability of 2 successes.} \\$
$p(getting \;doublet)=\frac{6}{36} \bigg[(1,1)(2,2) (3,3)(4,4) (5,5)(6,6)\bigg]\\$
$=\dfrac{1}{6}\\$
$q=\dfrac{5}{6}\\$
$n=4\\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3,...n \\$
$p(2)=4C_{2}\; \dfrac{1}{6}^{2}\;\dfrac{5}{6}^{2} \\$
$=6 \times \dfrac{1}{36}\; \dfrac{25}{36} \\$
$=\dfrac{150}{1296} \\$
$=\dfrac{25}{216} \\$

$\text{16. The mean of a binomial distribution is 5 and standard} \\$
$\text{deviation is 2. Determine the distribution.}\\$
$Mean=np=5 \\$
$SD=\sqrt{npq}=2 \\$
$npq=4 \\$
$5q=4 \\$
$q=\frac{4}{5} \\$
$p=1-q=1-\frac{4}{5}=\frac{1}{5} \\$
$Mean=np=5 \\$
$n\times\frac{1}{5}=5 \\$
$n=25 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3,...n \\$
$p(x)=25C_{x}\; \frac{1}{5}^{x}\;\frac{4}{5}^{25-x}\; x=0,1,2.3,...25 \\$

$\text{17. Determine the binomial distribution for which the mean}\; \\$
$\text{is 4 and variance 3. Also find}\; P(X=15) \\$
$Mean=np=4 \\$
$Var=npq=3 \\$
$npq=3 \\$
$4q=3 \\$
$q=\frac{3}{4} \\$
$p=1-q=1-\frac{3}{4}=\frac{1}{4} \\$
$Mean=np=4 \\$
$n\times\frac{1}{4}=4 \\$
$n=16 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3,...n \\$
$p(x=15)=16C_{15}\; (\frac{1}{4})^{15}\;\frac{3}{4}^{1} \\$
$=16 \times \dfrac{1}{4}^{15}\dfrac{3}{4}\\$
$=\dfrac{3}{4^{14}}\\$

$\text{18. Assume that a drug causes a serious side effect at a rate of three } \\$
$\text{patients per onehundred. What is the probability that atleast one person will } \\$
$\text{have side effects in a random sample of ten patients taking the drug?} \\$
$p=\frac{3}{100}=0.03 \\$
$q=1-0.03=0.97 \\$
$n=10 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3,...n \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3,...n \\$
$p(X) \ge 1=1-P(X < 1)=1-p(0) \\$
$1=0C_{0}(0.03)^{0}(0.97)^{10} \\$
$1-(1)(1)(0.7374) \\$
$p(X) \ge 1=0.2626 \\$

$\text{19. Consider five mice from the same litter, all suffering from Vitamin A }.\\$
$\text{deficiency They are fed a certain dose of carrots. The positive reaction } \\$
$\text{means recovery from the disease. Assume that the probability of recovery } \\$
$\text{is 0.73. What is the probability that atleast 3 of the 5 mice recover.} \\$
$p(recovery)=0.73 \\$
$q=1-0.73=0.27 \\$
$n=3 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3,...n \\$
$p(x \ge 3)= p(3)+p(4)+p(5)\\$
$5C_{3}\; (0.73)^{3}\;(0.27)^{2} +5C_{4}\; (0.73)^{4}\;(0.27)^{1}+\\$
$5C_{5}\; (0.73)^{5}\;(0.27)^{0}\\$
$=10(0.389017)(0.0729)+5(0.28398241)(0.27)+1(0.207307159)(1)\\$
$0.28359+0.383376+0.207307159\\$
$=0.874273159\\$
$p(x \ge 3)=0.8743 \\$

$\text{20. An experiment succeeds twice as often as it fails, what is the } \\$
$\text{probability that in next five trials there will be (i) three } \\$
$\text{successes and (ii) at least three successes.} \\$
$p(failing)=q\\$
$p (success)=p \\$
$Given p=2q \\$
$p=2(1-p) \\$
$p=2-2p \\$
$3p=2 \\$
$p=\frac{2}{3} \\$
$q=1-\frac{2}{3}=\frac{1}{3} \\$
$n=5 \\$
$p(x)=nC_{x}\; p^{x}\;q^{n-x}\; n=0,1,2.3,...n \\$

$(i)\; p(x= 3)= 5C_{3}\; \left( \dfrac{2}{3} \right)^{3}\;\left( \dfrac{1}{3} \right)^{2}\\$
$=10 \left( \dfrac{2}{3}\right)^{3}\; \left(\dfrac{1}{3}\right)^{2}\\$
$=\dfrac{80}{3^{5}} \\$
$p(x= 3)=\dfrac{80}{243} \\$

$(ii)(p \ge 3)=p(3)+p(4)+p(5) \\$
$5C_{3}\left( \dfrac{2}{3}\right)^{3}\;\left( \dfrac{1}{3}^{2}\right) \\$
$+ 5C_{4}\left(\dfrac{2}{3}\right)^{4}\;\left( \dfrac{1}{3}^{1}\right) \\$
$+5C_{5}\left(\dfrac{2}{3}\right)^{5}\;\left( \dfrac{1}{3}^{0}\right) \\$
$=10 \times \dfrac{2^{3}}{3^{5}}+5 \times \dfrac{2^{4}}{3^{5}}+(1)\dfrac{2^{5}}{3^{5}} \\$
$=\dfrac{80}{243}+\dfrac{80}{243}+\dfrac{32}{243} \\$
$=\dfrac{80+80+32}{243} \\$
$=(p \ge 3)=\dfrac{192}{243} \\$

Probability Distributions Exercise 7.1

Exercise 7.1

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