Solve the Following Question.(3 Marks)

Question 13 Marks

If the probability of success in a single trial is $0.01.$ How many trials are required in order to have a probability greater than $0.5$ of getting at least one success?

Answer

Let $X=$ number of successes. $p=$ probability of success in a single trial $\therefore \mathrm{p}=0.01$ and $q=1-p=1-0.01=0.99$ $\therefore \mathrm{X} \sim \mathrm{B}(\mathrm{n}, 0.01)$ The p.m.f. of $X$ is given by
$ \mathrm{P}(\mathrm{X}=\mathrm{x})={ }^n \mathrm{C}_x p^x q^{n-x}$
$\text { i.e. } p(x)={ }^n \mathrm{C}_x(0.01)^x(0.99)^{n-x}, x=1,2, \ldots, n$
$P(\text { at least one success })$
$\quad=P(X \geqslant 1)=1-P(X<1)$
$=1-P(X=0)=1-p(0)$
$=1-{ }^n \mathrm{C}_0(0.01)^0(0.99)^{n-0}$
$=1-1(1)(0.99)^n$
$=1-(0.99)^n $
Given: $P(X \geqslant 1)>0.5$
i.e. $1-(0.99)^n>0.5$
i.e. $1-0.5>(0.99)^n$
i.e. $0.5>(0.99)^n$
i.e. $(0.99)^n<0.5$
i.e. $\log (0.99)^n<\log (0.5)$
i.e. $n \log (0.99)<\log 0.5$
i.e. $n<\frac{\log 0.5}{\log 0.99}$
i.e. $n<68.96$
$\therefore n=68$
Hence, the number of trials required in order to have a probability greater than 0.5 of getting at least one success is $\frac{\log 0.5}{\log 0.99}$ or 68 .

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Question 23 Marks

The probability that a machine develops a fault within the first $3$ years of use is $0.003.$ If $40$ machines are selected at random, calculate the probability that $38$ or more will develop any faults within the first $3$ years of use.

Answer

Let $X=$ the number of machines who develop a fault.
$p=$ probability that a machine develops a fait within the first $3$ years of use
$\therefore p=0.003 \text { and } q=1-p=1-0.003=0.997$
Given: $n=40$
$\therefore \mathrm{X} \sim \mathrm{B}(40,0.003)$
The p.m.f. of $X$ is given by
$ P(X=x)={ }^n C_x p^x q^{n-x}, x=0,1,2, \ldots, n$
$\text { i.e. } p(x)={ }^{40} C_x(0.003)^x(0.997)^{40-x}, x=0,1,2, \ldots, 40 $
$P(38$ or more machines will develop any fault)
$ =P(X \geqslant 38)=P(X=38)+P(X=39)+P(X=40)$
$=p(38)+p(39)+p(40)$
$={ }^{40} \mathrm{C}_{38}(0.003)^{38}(0.997)^{40-38}+{ }^{40} \mathrm{C}_{39}(0.003)^{39}(0.997)^{40-39}+$
${ }^{40} \mathrm{C}_{40}(0.003)^{40}(0.997)^0$
$=\frac{40 \times 39}{2 \times 1}(0.003)^{38}(0.997)^2+40(0.003)^{39}(0.997)^1+$
$=(780)(0.003)^{38}(0.997)^2+(40)(0.003)^{39}(0.997)+$
$1 \cdot(0.003)^{40}(0.997)^0$
$=(0.003)^{38}\left[(780)(0.997)^2+40(0.003)(0.997)+(0.003)^2\right]$
$=(0.003)^{38}[775.327+0.1196+0.000009]$
$=(0.003)^{38}(775.446609)$
$=(775.446609)(0.003)^{38}$
$\approx(775.44)(0.003)^{38} $
Hence, the probability that $38$ or more machines will develop the fault within $3$ years of use $=(775.44)(0.003)^{38}$

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Question 33 Marks

The probability that a machine will produce all bolts in a production run within specification is 0.998. A sample of 8 machines is taken at random. Calculate the probability that (i) all 8 machines (ii) 7 or 8 machines (iii) at most 6 machines will produce all bolts within specification.

Answer

Let $\mathrm{X}=$ number of machines which produce the bolts within specification.
$p=$ probability that a machine produce bolts within specification
$p=0.998$ and $q=1-p=1-0.998=0.002$
Given: $\mathrm{n}=8$
$\therefore \mathrm{X} \sim \mathrm{B}(8,0.998)$
The p.m.f. of $X$ is given by
$ P(X=x)={ }^n C_x p^x q^{n-x}$
$\text { i.e. } \mathrm{p}(\mathrm{x})={ }^8 \mathrm{C}_x(0.998)^x(0.002)^{8-x}, \mathrm{x}=0,1,2, \ldots, 8 $
(i) $\mathrm{P}$ (all 8 machines will produce all bolts within specification) $=P[X=8]$
$ =p(8)$
$={ }^8 \mathrm{C}_8(0.998)^8(0.002)^{8-8}$
$=1(0.998)^8 \cdot(1)$
$=(0.998)^8 $
Hence, the probability that all 8 machines produce all bolts with specification $=$ $(0.998)^2$.
(ii) $P(7$ or 8 machines will produce all bolts within i specification $)=P(X=7)+P(X=8)$
$ ={ }^8 \mathrm{C}_7(0.998)^7(0.002)^{8-7}+{ }^8 \mathrm{C}_8(0.998)^8(0.002)^{8-8}$
$=8 \times(0.998)^7(0.002)^1+1 \times(0.998)^8(0.002)^0$
$=(0.998)^7[8(0.002)+0.998]$
$=(0.016+0.998)(0.998)^7$
$=(1.014) \times(0.998)^7 $
Hence, the probability that 7 or 8 machines produce all bolts within specification $=$ $(1.014)(0.998)^7$
$ \text { (iii) } P(a t \text { most } 6 \text { machines will produce all bolts with specification })=P[X \leq 6]$
$=1-P[X>6]$
$=1-[P(X=7)+P(X=8)]$
$=1-[P(7)+P(8)]$
$=1-(1.014)(0.998)^7 $
Hence, the probability that at most 6 machines will produce all bolts with specification $=1-(1.014)(0.998)^7$.

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Question 43 Marks

An examination consists of $10$ multiple choice questions, in each of which a candidate has to deduce which one of five suggested answers is correct. A completely unprepared student guesses each answer completely randomly. What is the probability that this student gets $8$ or more questions correct? Draw the appropriate moral.

Answer

Let $\mathrm{X}=$ number of correct answers.
$\mathrm{p}=$ probability that student gets correct answer
$ \therefore p=\frac{1}{5}$
$\therefore q=1-p=1-\frac{1}{5}=\frac{4}{5} $
Given: $n=10$ (number of total questions)
$\therefore X \sim B\left(10, \frac{1}{5}\right)$
The p.m.f. of $X$ is given by
$P(X=x)={ }^n \mathrm{C}_x p^x-q^{n-x}$
i.e. $p(x)={ }^{10} \mathrm{C}_x\left(\frac{1}{5}\right)^x\left(\frac{4}{5}\right)^{10-x}, x=0,1,2, \ldots, 10$
$P$ (student gets $8$ or more questions correct)
$= P(X \geqslant 8)=P(X=8)+P(X=9)+P(X=10)$ $ ={ }^{10} C_8\left(\frac{1}{5}\right)^8\left(\frac{4}{5}\right)^2+{ }^{10} C_9\left(\frac{1}{5}\right)^9\left(\frac{4}{5}\right)^1+{ }^{10} C_{10}\left(\frac{1}{5}\right)^{10}\left(\frac{4}{5}\right)^0$
$= \frac{10 \times 9 \times 8 !}{8 ! \times 2 \times 1} \times\left(\frac{1}{5}\right)^8 \times\left(\frac{4}{5}\right)^2+10\left(\frac{1}{5}\right)^9\left(\frac{4}{5}\right)^1+1 \times\left(\frac{1}{5}\right)^{10}$
$= 45 \times\left(\frac{1}{5}\right)^8 \times\left(\frac{4}{5}\right)^2+10 \times\left(\frac{1}{5}\right)^9 \times\left(\frac{4}{5}\right)+\left(\frac{1}{5}\right)^{10}$
$= \left(\frac{1}{5}\right)^8\left[45 \times\left(\frac{4}{5}\right)^2+10 \times\left(\frac{1}{5}\right) \times\left(\frac{4}{5}\right)+\left(\frac{1}{5}\right)^2\right]$
$= {\left[45 \times \frac{16}{25}+\frac{10}{5} \times \frac{4}{5}+\frac{1}{25}\right]\left(\frac{1}{5}\right)^8 }$
$= {\left[\frac{720}{25}+\frac{40}{25}+\frac{1}{25}\right]\left(\frac{1}{5^8}\right) }$
$= \left(\frac{761}{25}\right) \times\left(\frac{1}{5^8}\right)=\frac{30.44}{5^8}$
Hence, the probability that student gets $8$ or more questions correct $=\frac{30.44}{5^8}$

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Question 53 Marks

A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is $3\%.$ The inspector of the retailer picks $20$ items from a shipment. What is the probability that the store will receive at most one defective item?

Answer

Let $\mathrm{X}=$ number of defective electronic devices.
$\mathrm{p}=$ probability that device is defective
$ \therefore p=3 \%=\frac{3}{100}$
$\therefore q=1-p=1-\frac{3}{100}=\frac{97}{100} $
Given: $n=20$
$\therefore X \sim B\left(20, \frac{3}{100}\right)$
The p.m.f. of $X$ is given as:
$ P[X=x]={ }^n C_x p^x q^{n-x}$
$\text { i.e. } p(x)={ }^{20} C_x\left(\frac{3}{100}\right)^x\left(\frac{97}{100}\right)^{20-x} $
$P$ (store will receive at most one defective item)
$ =P[X \leqslant 1]=P[X=0]+P[X=1]$
$=p(0)+p(1)$
$={ }^{20} \mathrm{C}_0\left(\frac{3}{100}\right)^0\left(\frac{97}{100}\right)^{20-0}+{ }^{20} \mathrm{C}_1\left(\frac{3}{100}\right)^1\left(\frac{97}{100}\right)^{20-1}$
$=1 \times 1 \times(0.97)^{20}+20 \times(0.03) \times(0.97)^{19}$
$=(0.97+0.6)(0.97)^{19}$
$=(1.57)(0.97)^{19} $
Hence, the probability that the store will receive at most one defective item $=(1.57)$ $(0.97)^{19}$

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Question 63 Marks

The probability that a lamp in a classroom will be burnt out is $0.3.$ Six such lamps are fitted in the classroom. If it is known that the classroom is unusable if the number of lamps burning in it is less than four, find the probability that the classroom cannot be used on a random occasion.

Answer

Let $\mathrm{X}=$ number of lamps burnt out in the classroom.
$\mathrm{p}=$ probability of a lamp in a classroom will be burnt
$ \therefore p=0.3=\frac{3}{10}$
$\therefore q=1-p=1-\frac{3}{10}=\frac{7}{10} $
Given: $n=6$
$\therefore \mathrm{X} \sim \mathrm{B}\left(6, \frac{3}{10}\right)$
The p.m.f. of $X$ is given as:
$ \mathrm{P}[\mathrm{X}=\mathrm{x}]={ }^n \mathrm{C}_x p^x q^{n-x}$
$\text { i.e., } \mathrm{p}(\mathrm{x})={ }^6 \mathrm{C}_x\left(\frac{3}{10}\right)^x\left(\frac{7}{10}\right)^{6-x} $
Since the classroom is unusable if the number of lamps burning in it is less than four, therefore
$ P(\text { classroom cannot be used })=P[X<4]$
$=P[X=0]+P[X=1]+P[X=2]+P[X=3]$
$=p(0)+p(1)+p(2)+p(3)$
$={ }^6 \mathrm{C}_0\left(\frac{3}{10}\right)^0\left(\frac{7}{10}\right)^{6-0}+{ }^6 \mathrm{C}_1\left(\frac{3}{10}\right)^1\left(\frac{7}{10}\right)^{6-1}+$
${ }^6 \mathrm{C}_2\left(\frac{3}{10}\right)^2\left(\frac{7}{10}\right)^{6-2}+{ }^6 \mathrm{C}_3\left(\frac{3}{10}\right)^3\left(\frac{7}{10}\right)^{6-3}$
$=1 \times 1 \times\left(\frac{7}{10}\right)^6+6\left(\frac{3}{10}\right)\left(\frac{7}{10}\right)^5+$
$\quad \frac{6 \times 5}{1 \times 2} \cdot\left(\frac{3}{10}\right)^2\left(\frac{7}{10}\right)^4+\frac{6 \times 5 \times 4}{1 \times 2 \times 3} \cdot\left(\frac{3}{10}\right)^3\left(\frac{7}{10}\right)^3$
$=\left[7^6+18 \times 7^5+15 \times 9 \times 7^4+20 \times 27 \times 7^3 \frac{1}{10^6}\right.$
$=\frac{117649+302526+324135+185220}{10^6}$
$=\frac{929530}{10^6}=0.92953 $
Hence, the probability that the classroom cannot be used on a random occasion is $0.92953.$

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Question 73 Marks

The probability that a bomb will hit a target is $0.8.$ Find the probability that out of $10$ bombs dropped, exactly $2$ will miss the target.

Answer

Let $X=$ the number of bombs hitting the target.
$\mathrm{p}=$ probability that bomb will hit the target
$ \therefore p=0.8=\frac{8}{10}=\frac{4}{5}$
$\therefore q=1-p=1-\frac{4}{5}=\frac{1}{5} $
Given: $\mathrm{n}=10$
$\therefore \mathrm{X} \sim \mathrm{B}\left(10, \frac{4}{5}\right)$
The p.m.f. of $X$ is given as:
$ \mathrm{P}[\mathrm{X}=\mathrm{x}]={ }^n \mathrm{C}_x p^x q^{n-x}$
$\text { i.e. } \mathrm{p}(\mathrm{x})={ }^{10} \mathrm{C}_x\left(\frac{4}{5}\right)^x\left(\frac{1}{5}\right)^{10-x} $
$\mathrm{P}$ (exactly 2 bombs will miss the target $)=\mathrm{P}$ (exactly 8 bombs will hit the target)
$ =\mathrm{P}[\mathrm{X}=8]$
$=\mathrm{p}(8)$
$={ }^{10} \mathrm{C}_8\left(\frac{4}{5}\right)^8\left(\frac{1}{5}\right)^{10-8}$
$={ }^{10} \mathrm{C}_2\left(\frac{4}{5}\right)^8\left(\frac{1}{5}\right)^2$
$\left.=\frac{10 \times 9}{1 \times 2} \times \frac{4^8}{5^{10}}=\frac{45 \times 4^8}{5^{10}}=45\left(\frac{2^{16}}{5^{10}}\right){ }^n \mathrm{C}_{n-x}\right] $
Hence, the probability that exactly 2 bombs will miss the target $=45\left(\frac{2^{16}}{5^{10}}\right)$

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Question 83 Marks

Solve the following:
Let $X \sim B(10, 0.2).$ Find
(i) $P(X = 1)$
(ii) $P(X \geq 1)$
(iii) $P(X \leq 8).$

Answer

$ X \sim B(10,0.2)$
$\therefore n=10, p=0.2$
$\therefore q=1-p=1-0.2=0.8 $
The p,m.f. of $X$ is given by
$ P(X=x)={ }^n \mathrm{C}_x p^x q^{n-x}$
$\therefore P(X=x)={ }^{10} \mathrm{C}_x(0.2)^x(0.8)^{10-x}, x=0,1,2,3, \ldots, 10$
$\text { (i) } P(X=1) ={ }^{10} \mathrm{C}_1(0.2)^1(0.8)^{10-1}$
$ =10(0.2)(0.8)^9=2(0.8)^9 .$
(ii) $ P(X \geqslant 1) =1-P(X<1)$
$ =1-P(X=0)$
$=1-{ }^{10} C_0(0.2)^0(0.8)^{10-0}$
$=1-1 \times 1 \times(0.8)^{10}=1-(0.8)^{10} $
(iii) $ P(X \leqslant 8) =1-P(X>8)$
$ =1-[P(X=9)+P(X=10)]$
$ =1-\left[{ }^{10} C_9(0.2)^9(0.8)^{10-9}+\right.$
$ \left.{ }^{10} C_{10}(0.2)^{10}(0.8)^{10-10}\right]$
$ =1-\left[10(0.2)^9(0.8)^1+1(0.2)^{10}(0.8)^0\right]$
$ =1-(0.2)^9[10(0.8)+(0.2)]$
$=1-(0.2)^9[8+0.2]$
$ =1-(8.2)(0.2)^9$

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Question 93 Marks

In binomial distribution with five Bernoulli’s trials, the probability of one and two success are 0.4096 and 0.2048 respectively. Find the probability of success.

Answer

Given: $X \sim B(n=5, p)$
The probability of $X$ success is
$
P(X=x)={ }^n \mathrm{C}_x p^x q^{n-x}, x=0,1,2, \ldots, n
$
i.e. $P(X=x)={ }^5 \mathrm{C}_x p^x q^{5-x}, x=0,1,2,3,4,5$
Probabilities of one and two successes are
$
P(X=1)={ }^5 C_1 p^1 q^{5-1}
$
and $P(X=2)={ }^5 \mathrm{C}_2 p^2 q^{5-2}$ respectively
Given: $P(X=1)=0.4096$ and $P(X=2)=0.2048$
$\therefore \frac{P(X=2)}{P(X=1)}=\frac{0.2048}{0.4096}$
i.e. $\frac{{ }^5 \mathrm{C}_2 p^2 q^{5-2}}{{ }^5 \mathrm{C}_1 p^2 q^{5-1}}=\frac{1}{2}$
i.e. $2 \times{ }^5 C_2 p^2 q^3=1 \times{ }^5 C_1 p q^4$
i.e. $2 \times \frac{5 \times 4}{1 \times 2} \times p^2 q^3=1 \times 5 \times p q^4$
i.e. $20 p^2 q^3=5 p q^4$
i.e. $4 p=q$
i.e. $4 p=1-p$
i.e. $5 p=1$
$
\therefore p=\frac{1}{5}
$
Hence, the probability of success is $\frac{1}{5}$.

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Question 103 Marks

If a fair coin is tossed 10 times, find the probability of getting
(i) exactly six heads
(ii) at least six heads
(iii) at most six heads

Answer

The repeated tosses of a coin are Bernoulli trials. Let $X$ denote the number of heads in an experiment of 10 trials.
Clearly, $X \sim B(n, p)$ with $n=10$ and $p=\frac{1}{2}, q=1-p=1-\frac{1}{2} \quad \therefore \quad q=\frac{1}{2}$
$
\begin{aligned}
P(X=x) & ={ }^n C_x p^x \times q^{n-x} \\
& ={ }^{10} C_x\left(\frac{1}{2}\right)^x \times\left(\frac{1}{2}\right)^{n-x}
\end{aligned}
$
(i) Exactly six successes means $x=6$
$
\begin{aligned}
P(X=6)={ }^{10} C_6\left(\frac{1}{2}\right)^6 \times\left(\frac{1}{2}\right)^{10-6}=\frac{10 !}{6 !(10-6) !} \times\left(\frac{1}{2}\right)^6 \times\left(\frac{1}{2}\right)^4 & =\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times\left(\frac{1}{2}\right)^{10} \\
& =\frac{105}{512}
\end{aligned}
$
(ii) At least six successes means $x \geq 6$
$
\begin{aligned}
& P(X \geq 6)=[P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)] \\
& ={ }^{10} C_6\left(\frac{1}{2}\right)^6 \times\left(\frac{1}{2}\right)^4+{ }^{10} C_7\left(\frac{1}{2}\right)^7 \times\left(\frac{1}{2}\right)^3+{ }^{10} C_8\left(\frac{1}{2}\right)^8 \times\left(\frac{1}{2}\right)^2+{ }^{10} C_9\left(\frac{1}{2}\right)^9 \times\left(\frac{1}{2}\right)^1+{ }^{10} C_{10}\left(\frac{1}{2}\right)^{10} \times\left(\frac{1}{2}\right)^0 \\
& =\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times\left(\frac{1}{2}\right)^{10}+\frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times\left(\frac{1}{2}\right)^{10}+\frac{10 \times 9}{2 \times 1} \times\left(\frac{1}{2}\right)^{10}+10\left(\frac{1}{2}\right)^{10}+\left(\frac{1}{2}\right)^{10} \\
& =(210+120+45+10+1) \times \frac{1}{1024} \\
& =\frac{386}{1024}=\frac{193}{512}
\end{aligned}
$
(iii) At most six successes means $x \leq 6$
$
\begin{aligned}
& P(X \leq 6)=1-(\mathrm{P}(X>6) \\
& =1-[P(X=7)+P(X=8)+P(X=9)+P(X=10)] \\
& =1-\left[{ }^{10} C_7\left(\frac{1}{2}\right)^7 \times\left(\frac{1}{2}\right)^3+{ }^{10} C_8\left(\frac{1}{2}\right)^8 \times\left(\frac{1}{2}\right)^2+{ }^{10} C_9\left(\frac{1}{2}\right)^9 \times\left(\frac{1}{2}\right)^1+{ }^{10} C_{10}\left(\frac{1}{2}\right)^{10} \times\left(\frac{1}{2}\right)^0\right] \\
& =1-\left[\frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times\left(\frac{1}{2}\right)^{10}+\frac{10 \times 9}{2 \times 1} \times\left(\frac{1}{2}\right)^{10}+10\left(\frac{1}{2}\right)^{10} \times\left(\frac{1}{2}\right)^{10}\right] \\
& =1-\left[(120+45+10+1) \times \frac{1}{1024}\right]=1-\frac{176}{1024}=1-\frac{88}{512}=\frac{512-88}{512}=\frac{424}{512}=\frac{53}{64}
\end{aligned}
$

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Question 113 Marks

It is known that $10\%$ of certain articles manufactured are defective. What is the probability that in a random sample of $12$ such articles, $9$ are defective?

Answer

Let $X=$ number of defective articles.
$p=$ probability of defective articles.
$\therefore p=10 \%=\frac{10}{100}=\frac{1}{10}$
and $q=1-p=1-\frac{1}{10}=\frac{9}{10}$
Given: $n=12$
$\therefore X \sim B\left(12, \frac{1}{10}\right)$
The p.m.f. of $X$ is given by
$ P[X=x]={ }^n \mathrm{C}_x p^x \mathrm{q}^{n-x}$
$\text { i.e. } p(x)={ }^{12} \mathrm{C}_x\left(\frac{1}{10}\right)^x\left(\frac{9}{10}\right)^{12-x}, x=1,2,3, \ldots, 12 $
i.e. $p(x)={ }^{12} C_x\left(\frac{1}{10}\right)^x\left(\frac{9}{10}\right)^{12-x}, x=1,2,3, \ldots, 12$
$P(9$ defective articles $)=P[X=9]$
$=p(9)={ }^{12} \mathrm{C}_9\left(\frac{1}{10}\right)^9\left(\frac{9}{10}\right)^{12-9}$
$=\frac{12 !}{9 ! 3 !}\left(\frac{1}{10}\right)^9\left(\frac{9}{10}\right)^3$
$=\frac{12 \times 11 \times 10 \times 9 !}{9 ! \times 3 \times 2 \times 1} \times \frac{1}{10^9} \times \frac{9^3}{10^3}$
$=2 \times 11 \times 10 \cdot \frac{9^3}{10^{12}}=22\left(\frac{9^3}{10^{11}}\right)$
Hence, the probability of getting 9 defective articles $=22\left(\frac{9^3}{10^{11}}\right)$

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Question 123 Marks

A person buys a lottery ticket in $50$ lotteries, in each of which his chance of winning a prize is $\frac{1}{100}$, find the probability that he will win a prize
(i) at least once
(ii) exactly once
(iii) at least twice.

Answer

Let $X=$ number of winning prizes.
$p=$ probability of winning a prize
$\therefore p=\frac{1}{100}$
and $q=1-p=1-\frac{1}{100}=\frac{99}{100}$
Given: $n=50$
$\therefore \mathrm{X} \sim \mathrm{B}\left(50, \frac{1}{100}\right)$
The p.m.f. of $X$ is given by
$ P(X=x)={ }^n C_x p^x q^{n-x}$
$\text { i.e., } p(x)={ }^{50} C_x\left(\frac{1}{100}\right)^x\left(\frac{99}{100}\right)^{50-x}, x=0,1,2, \ldots 50 $
(i) $P$ (a person wins a prize at least once)
$ =P[X \geqslant 1]=1-P[X<1]=1-p(0)$
$=1-{ }^{50} \mathrm{C}_0\left(\frac{1}{100}\right)^0\left(\frac{99}{100}\right)^{50-0} $
$ =1-1 \times 1 \times\left(\frac{99}{100}\right)^{50}$
$=1-\left(\frac{99}{100}\right)^{50} $
Hence, probability of winning a prize at least once $=1-\left(\frac{99}{100}\right)^{50}$
(ii) $P$ (a person wins exactly one prize $)=P[X=1]=p(1)$
$ ={ }^{50} \mathrm{C}_1\left(\frac{1}{100}\right)^1\left(\frac{99}{100}\right)^{50-1}$
$=50 \times\left(\frac{1}{100}\right) \times\left(\frac{99}{100}\right)^{49}$
$=\frac{1}{2}\left(\frac{99}{100}\right)^{49} $
Hence, probability of winning a prize exactly once $=\frac{1}{2}\left(\frac{99}{100}\right)^{49}$
$ \text { (iii) } P(a \text { persons wins the prize at least twice })=P[X \geq 2]$
$=1-P[X<2]$
$=1-[p(0)+p(1)]$
$=1-\left[{ }^{50} C_0\left(\frac{1}{100}\right)^0\left(\frac{99}{100}\right)^{50-0}+\right.$
$\left.{ }^{50} C_1\left(\frac{1}{100}\right)^1\left(\frac{99}{100}\right)^{50-1}\right] $
$\left.{ }^{50} \mathrm{C}_1\left(\frac{1}{100}\right)^{\prime}\left(\frac{99}{100}\right)^{50-1}\right]$
$ =1-\left[1 \times 1 \times\left(\frac{99}{100}\right)^{50}+50 \times \frac{1}{100} \times\left(\frac{99}{100}\right)^{49}\right]$
$=1-\left[\left(\frac{99}{100}\right)^{50}+\frac{1}{2}\left(\frac{99}{100}\right)^{49}\right]$
$=1-\left(\frac{99}{100}\right)^{49}\left[\frac{99}{100}+\frac{1}{2}\right]$
$=1-\left(\frac{99}{100}\right)^{49}\left[\frac{149}{100}\right]$
$=1-149\left(\frac{99^{40}}{100^{50}}\right) $
Hence, the probability of winning the prize at least twice $=1-149\left(\frac{99^{49}}{100^{50}}\right)$.

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