4.21.8.1 Solved Examples

Example 4.54: There are 50 telephone lines in an exchange. The probability of them will be busy is 0.1. What is the probability that all the lines are busy?

Solution: Let X be the Poisson variable “number of busy line in the exchange”

By Poisson distribution $P(X=x)=\frac{{\lambda }^x{e}^{-x}}{x!},x=0,1,2,3,\dots$

We have n=50, p=0.1

$\lambda =np=(50)(0.1)=5$

Hence P(all lines are busy)=$P(X=50)=\frac{5^{50}{e}^{-5}}{50!}$

Example 4.55: The probability that a bomb dropped from an envelope will strike a certain target is $1/5$. If 6 bombs are dropped, find the probability that

Given e^−1.2=0.3012

Solution: We have n=6, $p=1/5$

$\lambda =np=(6)\left(\frac{1}{5}\right)=1.2$

By Poisson distribution $P(X=x)=\frac{{\lambda }^x{e}^{-\lambda }}{x!},x=0,1,2,3,\dots$

Example 4.56: If the probability that an individual suffers a bad reaction from an injection of a given serum is 0.001. Determine the probability that out of 2000 individuals exactly 3 individuals will suffer due to bad reaction.

Solution: We have

$\begin{array}{l}n=2000,p=0.001(\mathrm{given})\hfill \\ \lambda =np=(2000)(0.001)=2\hfill \end{array}$

By Poisson distribution

$P(X=x)=\frac{{\lambda }^x{e}^{-\lambda }}{x!},x=0,1,2,3,\dots$

P(X=3)=Probability that 3 individuals will suffer due to bad reaction $\frac{2^3{e}^{-2}}{3!}=0.1805$

Example 4.57: For a Poisson distribution show that ${\mu }_{r+1}=r\lambda {\mu }_{r-1}+\lambda \frac{d{\mu }_r}{d\lambda }$ where λ is the mean.

Solution: The rth moment of the distribution is

$\begin{array}{ll}{\mu }_r\hfill & =\sum _{x=0}^{\infty }{(x-\lambda )}^{r}P(X=x)\hfill \\ \hfill & =\sum _{x=0}^{\infty }{(x-\lambda )}^r\frac{{\lambda }^x{e}^{-1}}{x!}\hfill \end{array}$

Differentiating with respect to λ we get

$\begin{array}{ll}\frac{\mathrm{d}{\mu }_r}{\mathrm{d}\lambda }\hfill & =–r\sum _{x=0}^{\infty }{(x-\lambda )}^{r-1}\frac{e^{-\lambda }{\lambda }^x}{x!}+\sum _{x=0}^{\infty }{(x-\lambda )}^r\left[\frac{x{\lambda }^{n-1}{e}^{-\lambda }-{\lambda }^x{e}^{-\lambda }}{x!}\right]\hfill \\ \hfill & =–r{\mu }_{r-1}+\sum _{x=0}^{\infty }{(x-\lambda )}^r\left[\frac{{\lambda }^{n-1}{e}^{-\lambda }(x-\lambda )}{x!}\right]\hfill \end{array}$

$\begin{array}{ll}\therefore \lambda \frac{\mathrm{d}{\mu }_r}{\mathrm{d}\lambda }\hfill & =–r\lambda {\mu }_{r-1}+\sum _{x=0}^{\infty }\left[\frac{{\lambda }^n{e}^{-\lambda }{\left(x–\lambda \right)}^{r+1}}{x!}\right]\hfill \\ \hfill & =–\lambda {\mu }_{r–1}+{\mu }_{r+1}\hfill \end{array}$

Hence

$\lambda \frac{\mathrm{d}{\mu }_r}{\mathrm{d}\lambda }+r\lambda {\mu }_{r-1}={\mu }_{r+1}$

i.e.,

${\mu }_{r+1}=r\lambda {\mu }_{r-1}+\lambda \frac{\mathrm{d}{\mu }_r}{\mathrm{d}\lambda }$

Example 4.58: Suppose that $P(X=2)=\frac{2}{3}P(X=1)$, find P(X=0)?

Solution:

$P(X=2)=\frac{2}{3}P(X=1)(\mathrm{given})$

$\frac{{\lambda }^2{e}^{–\lambda }}{2!}=\frac{2}{3}\frac{{\lambda }^1{e}^{–\lambda }}{1!}$

or

$\frac{{\lambda }^2}{2!}=\frac{2}{3}\lambda$

or

$3{\lambda }^2–4\lambda =0$

or

$\lambda (3\lambda –4)=0$

$\lambda =0\mathrm{or}\lambda =\frac{4}{3}$

We consider $\lambda =4/3$, since λ=0 is impossible

$P(X=0)=\frac{{\lambda }^0{e}^{–\lambda }}{0!}=e^{–4/3}.$

Example 4.59: A car hire firm has two cars, which it hires out day by day. The number of demands for a car on each day is distributed as Poisson distribution with mean 1.5. Calculate the proportion of days on which neither car is used and the proportion of days on which some demand is refused (e^−1.5=0.2231).

Solution: We have λ=1.5 (given)

The proportion of days when no car will be required is

$P(X=0)=\frac{{\lambda }^0{e}^{–\lambda }}{0!}+e^{–\lambda }=e^{–1.5}=0.2231$

The probability that no car, one car, two cars will be required

$\begin{array}{l}=P(X=0)+P(X=1)+P(X=2)\hfill \\ =e^{–\lambda }+e^{–\lambda }\frac{\lambda }{1!}+e^{–\lambda }\frac{{\lambda }^2}{2!}\hfill \\ =e^{–\lambda }\left(1+\frac{\lambda }{1!}+\frac{{\lambda }^2}{2!}\right)\hfill \\ =e^{–1.5}(1+1.5+1.125)\hfill \\ =(0.2231)(3.625)\hfill \\ =0.80873\hfill \end{array}$

Example 4.60: In a Poisson distribution P(X=x) for x=10 is 10%. Find the mean, given that log_e10=2.3026.

Solution: Let λ be the mean

$\begin{array}{l}P(X=x)=P(X=0)=\frac{{\lambda }^0{e}^{-\lambda }}{0!}=e^{-\lambda }=\frac{10}{100}\hfill \\ \therefore e^{-\lambda }=0.1=\frac{1}{10}\mathrm{or}\therefore e^{\lambda }=10\hfill \end{array}$

Hence

$\lambda ={\log }_{e}10$

Example 4.61: At a busy traffic intersection, the probability p of an individual car having an accident is very small, say, p=0.0001. However, during a certain part of the day, a large number of cars, say 1000 pass through the intersection. Under these conditions, what is the probability of two or more accidents occurring during that period?

Solution: Here we have n=1000, p=0.0001

$\begin{array}{ll}\therefore \lambda \hfill & =np=1000\times 0.0001=0.1\hfill \\ P(X\ge 2)\hfill & =1–P(X<2)=1–[P(X=0)+P(X=1)]\hfill \\ \hfill & =1-\left[\frac{e^{-0.1}{(0.1)}^0}{0!}+\frac{e^{-0.1}{(0.1)}^1}{1!}\right]\hfill \\ \hfill & =1–e^{–0.1}(1+0.1)\hfill \\ \hfill & =1–(0.9048)(1.1)\hfill \\ \hfill & =0.00472\hfill \end{array}$

Example 4.62: Using Poisson distribution, find the probability that the ace of spades will be drawn from a pack of well shuffled cards at least once in 104 consecutive trials.

Solution: Probability that the card drawn is an ace of spades=$1/52$.

Since n=104

Mean of the distribution=$\lambda =np=104\times \frac{1}{52}=2$

Probability of getting ace of spades at least once=P(X≥1)=1−P(X=0)

$\begin{array}{l}=1-\frac{{\lambda }^0{e}^{-\lambda }}{0!}=1-e^{-2}\hfill \\ =1–0.136\hfill \\ =0.864\hfill \end{array}$

Example 4.63: Probability of getting no misprint in a page of a book is e⁻⁴. What is the probability that a page contains more than two misprints?

Solution: It is given that P(X=0), i.e., probability of getting no misprint=e⁻⁴

i.e.,

$\therefore \frac{{\lambda }^0{e}^{-\lambda }}{0!}=e^{-4}$

or

$e^{-\lambda }=e^{-4}$

i.e.,

$\lambda =4$

Probability that a page contains more than 2 misprints=P(X>2)=1−P(X≤2).

$\begin{array}{l}=1–[P(X=0)+P(X=1)+P(X=2)]\hfill \\ =1–\left[e^{–4}+e^{-4}0.4+e^{–4}\frac{4^2}{2!}\right]\hfill \\ =1–e^{–4}[1+4+8]\hfill \\ =0.7618\hfill \end{array}$

Example 4.64: Six coins are tossed 6400 times using Poisson distribution, what is the approximate probability of getting six heads 10 times?

Solution: Probability of getting a head when a coin is tossed=$\frac{1}{2}$

Probability of getting 6 heads when 6 coins are tossed=${\left(\frac{1}{2}\right)}^6=\frac{1}{64}$

Mean=$\lambda =np=6400\times \frac{1}{64}=100$

Probability of getting 6 heads 10 times=$P(X=10)=\frac{{\lambda }^{10}{e}^{-\lambda }}{10!}=\frac{e^{-100}{(100)}^{10}}{100!}$.

Example 4.65: Suppose 2% of the people on the average are left handed. Find (1) the probability of finding 3 or more left handed; (2) the probability of finding none or one left handed?

Solution: Let x be the random variable the number of left handed people.

The mean is=$\lambda =2\%=\frac{2}{100}=0.02$

Since $P(X=x)=\frac{e^{-\lambda }{\lambda }^x}{x!},x=1,2,3,\dots$

Example 4.66: If X is a Poisson variate such that $3P(X=4)=\frac{1}{2}P(X=2)+P(X=0)$

Find

Solution:

1. Since X is a Poisson variate, we have

$P(X=x)=\frac{e^{-\lambda }{\lambda }^x}{x!},x=1,2,3,\dots$

Consider $3P(X=4)=\frac{1}{2}P(X=2)+P(X=0)$ (given)
or

$\frac{e^{-\lambda }{\lambda }^4}{4!}=\frac{1}{2}\frac{e^{-\lambda }{\lambda }^2}{2!}+\frac{e^{-\lambda }{\lambda }^0}{0!}$

or

$\frac{{\lambda }^4}{8}=\frac{{\lambda }^2}{4}+1$

or

${\lambda }^4-2{\lambda }^2-8=0$

or

$({\lambda }^4-4)({\lambda }^2+2)=0$

since λ>0, we get λ=2
Mean of the Poisson variate λ=2
i.e., $\lambda =2$

2. $\begin{array}{ll}P(X\le 2)\hfill & =P(X=0)+P(X=1)+P(X=2)\hfill \\ \hfill & =\frac{e^{-2}{2}^0}{0!}+\frac{e^{-2}{2}^1}{1!}+\frac{e^{-2}{2}^2}{2!}\hfill \\ \hfill & =0.675\hfill \end{array}$

Example 4.67: Average number of accidents on any day on a national highway is 1.8. Determine the probability that the number of accidents (1) is at least one; (2) at most one.

Solution: Average number of accidents=1.8

i.e., λ=mean=1.8

1. P(X≥1)=Probability that the number of accidents is at least one

$\begin{array}{l}=1–P(X=0)\hfill \\ =1–e^{–1.8}\hfill \\ =1–0.1653=0.8347\hfill \end{array}$

2. Probability that the number of accidents is at most one=P(X≤1)

$\begin{array}{l}=P(X=0)+P(X=1)\hfill \\ =\frac{e^{–1.8}{(1.8)}^0}{0!}+\frac{e^{–1.8}{(1.8)}^1}{1!}\hfill \\ =e^{–1.8}(1+1.8)\hfill \\ =0.4628\hfill \end{array}$

Example 4.68: Using the recurrence formula, find the probabilities when x=0, 1, 2, 3, 4, and 5, if the mean of the Poisson distribution is 3.

Solution: We have λ=3 (given)

$P(X=0)=P(0)=e^{–\lambda }=e^{–3}$

The recurrence formula is $P(x+1)=\frac{\lambda }{x+1}P(x)$

Substituting x=0, 1, 2, 3, 4, and 5, we get

$\begin{array}{l}P(0+1)=P(1)=\frac{\lambda }{0+1}P(0)=3(e^{–3})=0.147\hfill \\ P(1+1)=P(2)=\frac{\lambda }{1+1}P(1)=\frac{3}{2}(0.147)=0.2205)\hfill \\ P(2+1)=P(3)=\frac{\lambda }{2+1}P(2)=\frac{3}{3}(0.2205)=0.2205\hfill \\ P(3+1)=P(4)=\frac{\lambda }{3+1}P(3)=\frac{3}{4}(0.2205)=0.1653\hfill \\ P(4+1)=P(5)=\frac{\lambda }{4+1}P(4)=\frac{3}{5}(0.1653)=0.099\hfill \\ P(5+1)=P(6)=\frac{\lambda }{5+1}P(5)=\frac{3}{6}(0.099)=0.0495\hfill \\ P(1)=0.147,P(2)=0.2205P(3)=0.2205,\hfill \\ P(4)=0.1653,P(5)=0.099P(3)=0.0495\hfill \end{array}$

Example 4.69: Wireless sets are manufactured with 25 soldered joints each. On an average 1 joint in 500 is defective. How many sets can be expected to be free from defective joints in a consignment of 10,000 sets?

Solution: Probability that a joint is defective=$p=\frac{1}{500}$.

We have n=25, $p=\frac{1}{500}$

Mean=λ=np=$25\times \frac{1}{500}=0.05$

By Poisson distribution $P(X=x)=\frac{e^{–\lambda }{\lambda }^x}{x!},x=1,2,3,\dots$

Probability that no joint is defective=$P(0)=\frac{e^{–\lambda }{\lambda }^0}{0!}=e^{–0.05}$

Thus the expected number of sets free from defects in 10,000 sets

$\begin{array}{l}=10,000\times P(0)=10,000\times e^{-0.05}\hfill \\ =10,000(0.9512)\hfill \\ =9512.0\hfill \end{array}$

9512 Sets are expected to be free from defects.

Example 4.70: Fit a Poisson distribution to the following data and calculate the theoretical (expected) frequencies.

Solution: Mean of the distribution=$\frac{0.122+1.60+2.15+3.2+4.1}{122+60+15+2+1}$

$\begin{array}{l}\overline{x}=\frac{60+30+6+4}{200}=\frac{100}{200}\hfill \\ \therefore \lambda =\overline{x}=0.5\hfill \end{array}$

We have N=122+60+15+2+1=200

$N=4,\lambda =0.5$

$P(0)=P(X=0)=\frac{e^{–\lambda }{\lambda }^0}{0!}=e^{–0.5}$

The theoretical frequencies are obtained as follows:

$\begin{array}{ll}N.P(0)\hfill & =200\cdot e^{–0.5}=121\hfill \\ N.P(1)\hfill & =200\cdot \frac{e^{–0.5}{(0.5)}^1}{1!}=61\hfill \\ N.P(2)\hfill & =200\cdot \frac{e^{–0.5}{(0.5)}^2}{2!}=15\hfill \\ N.P(3)\hfill & =200\cdot \frac{e^{–0.5}{(0.5)}^3}{3!}=3\hfill \\ N.P(4)\hfill & =200\cdot \frac{e^{–0.5}{(0.5)}^4}{4!}=0\hfill \end{array}$

Hence the expected (theoretical) frequencies are 121, 61, 15, 3, 0.

Example 4.71: The variance of the Poisson distribution is 2. Find the distribution for x=0, 1, 2, 3, 4, and 5 from the recurrence relation of the distribution (e⁻²=0.1353)

Solution: The variance of the Poisson distribution λ=2 (given).

We have P(0)=e⁻²=0.1353 (given)

The recurrence relation is $P(x+1)=\frac{\lambda }{x+1}P(x)$

Hence we get

$\begin{array}{l}P(1)=\frac{\lambda }{0+1}P(0)=\frac{2}{1}{e}^{–2}=2\times 0.1353=0.2706\hfill \\ P(2)=\frac{\lambda }{1+1}P(1)=\frac{2}{2}(0.2706)=0.2706\hfill \\ P(3)=\frac{\lambda }{2+1}P(2)=\frac{2}{3}(0.2706)=0.1804\hfill \\ P(4)=\frac{\lambda }{3+1}P(3)=\frac{2}{4}(0.1804)=0.0902\hfill \\ P(5)=\frac{\lambda }{4+1}P(4)=\frac{2}{5}(0.0902)=0.361\hfill \end{array}$

Example 4.72: Assuming that the typing mistakes per page committed by a typist follows a Poisson distribution, find the theoretical frequencies for the following distribution of typing mistakes?

Solution: We have N=40+30+20+15+10+5=120, n=5

$\begin{array}{cc}\mathrm{Mean}\mathrm{of}\mathrm{the}\mathrm{distribution}=\lambda & =\frac{0.40+1.30+2.20+3.15+4.10+5.5}{120}\\ & =\frac{30+40+45+40+25}{120}=\frac{180}{120}=1.5\end{array}$

Using Poisson distribution we get

$\begin{array}{ll}P(0)\hfill & =\frac{e^{–\lambda }{\lambda }^0}{0!}=e^{–\lambda }=e^{–1.5}=0.22313\hfill \\ P(1)\hfill & =\frac{e^{–\lambda }{\lambda }^1}{1!}=e{.}^{–\lambda }\lambda =e^{–1.5}(1.5)=0.334695\hfill \\ P(2)\hfill & =\frac{e^{–\lambda }{\lambda }^2}{2!}=\frac{e^{–1.5}{(1.5)}^2}{2}=025\hfill \\ P(3)\hfill & =\frac{e^{–\lambda }{\lambda }^3}{3!}=\frac{e^{–1.5}{(1.5)}^3}{6}=0.13\hfill \\ P(4)\hfill & =\frac{e^{–\lambda }{\lambda }^4}{4!}=\frac{e^{–1.5}{(1.5)}^4}{24}=0.05\hfill \\ P(5)\hfill & =\frac{e^{–\lambda }{\lambda }^5}{5!}=\frac{e^{–1.5}{(1.5)}^5}{120}=0.01\hfill \end{array}$