Solve the Following Question.(2 Marks)

Question 12 Marks

The probability that a certain kind of component will survive a check test is $0.6.$ Find the probability that exactly $2$ of the next $4$ tested components tested survive.

Answer

Let $\mathrm{X}=$ number of tested components survive.
$p=$ probability that the component survives the check test
$ \therefore p=0.6=\frac{6}{10}=\frac{3}{5}$
$\therefore q=1-p=1-\frac{3}{5}=\frac{2}{5} $
Given : $n=4$
$\therefore X \sim B\left(4, \frac{3}{5}\right)$
The p.m.f. of $X$ is given as :
$P[X=x]={ }^n \mathrm{C}_x p^x q^{n-x}$
i.e. $p(x)={ }^4 C_x\left(\frac{3}{5}\right)^x\left(\frac{2}{5}\right)^{4-x}$
$P \text { (exactly } 2 \text { components survive) }$
$ =P[X=2]=p(2)$
$={ }^4 C_2\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^{4-2}$
$=\left(\frac{4 \times 3}{1 \times 2}\right) \times\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^2=\frac{6 \times 9 \times 4}{625}$
$=\frac{216}{625}=0.3456 $
Hence, the probability that exactly $2$ of the $4$ tested components survive is $0.3456 .$

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Question 22 Marks

A lot of $100$ items contain $10$ defective items. Five items are selected at random from the lot and sent to the retail store. What is the probability that the store will receive at most one defective item?

Answer

Let $\mathrm{X}=$ number of defective items.
$\mathrm{p}=$ probability that item is defective
$ \therefore p=\frac{10}{100}=\frac{1}{10}$
$\therefore q=1-p=1-\frac{1}{10}=\frac{9}{10} $
Given: $\mathrm{n}=5$
$\therefore \mathrm{X} \sim \mathrm{B}\left(5, \frac{1}{10}\right)$
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})={ }^5 C_x\left(\frac{1}{10}\right)^x\left(\frac{9}{10}\right)^{5-x} $
$ P(\text { store will receive at most one defective item })=P[X \leq 1]$
$=P[X=0]+P[X=1]$
$=P(0)+P(1)$
$={ }^5 C_0\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^{5-0}+{ }^5 C_1\left(\frac{1}{10}\right)^1\left(\frac{9}{10}\right)^{5-1}$
$=1 \times 1 \times\left(\frac{9}{10}\right)^5+5 \times \frac{1}{10} \times\left(\frac{9}{10}\right)^4$
$=(0.9)^5+(0.05)(0.9)^4$
$=(0.9+0.5)(0.9)^4$
$=(1.4)(0.9)^4 $
Hence, the probability that the store will receive at most one defective item is $(1.4) (0.9)^4$

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Question 32 Marks

Ten eggs are drawn successively with replacement from a lot containing $10 \%$ defective eggs. Find the probability that there is at least one defective egg.

Answer

Let $X$ denote the number of defective eggs in the 10 eggs drawn.
Since the drawing is done with replacement, the trials are Bernoulli trials.
$
\begin{aligned}
& \text { Probability of success }=\frac{1}{10} \\
& \qquad \begin{aligned}
& p=\frac{1}{10}, q=1-p=1-\frac{1}{10} \quad \therefore \quad q=\frac{9}{10} \\
& X \sim B\left(10, \frac{1}{10}\right) \\
& \quad n=10 \\
& \operatorname{Here} X \geq 1 \\
& P(X \geq 1)=1-{ }^{10} C_0\left(\frac{1}{10}\right)^0 \times\left(\frac{9}{10}\right)^{10} \\
&=1-1 \times 1 \times\left(\frac{9}{10}\right)^{10} \times\left(\frac{9}{10}\right)^{10-x} \\
&=1-\left(\frac{9}{10}\right)^{10}
\end{aligned}
\end{aligned}
$

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Question 42 Marks

Six balls are drawn successively from an urn containing 7 red and 9 black balls. Tell whether or not the trials of drawing balls are Bernoulli trials when after each draw the ball drawn is
(i) replaced
(ii) not replaced in the urn.

Answer

(i) The number of trials is finite. When the drawing is done with replacement, the probability of success (say, red ball) is $p=\frac{7}{16}$ which is same for all six trials (draws). Hence, the drawing of balls with replacements are Bernoulli trials.
(ii) When the drawing is done without replacement, the probability of success (i.e. red ball) in first trial is $\frac{7}{16}$ in second trial is $\frac{6}{15}$ if first ball drawn is red and is $\frac{7}{15}$ if first ball drawn is black and so on. Clearly probability of success is not same for all trials, hence the trials are not Bernoulli trials.

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Question 52 Marks

On a multiple-choice examination with three possible answers for each of the five questions. What is the probability that a candidate would get four or more correct answers just by guessing?

Answer

Let $X=$ number of correct answers.
$p=$ probability that a candidate gets a correct answer from three possible answers.
$\therefore p=\frac{1}{3}$ and $q=1-p=1-\frac{1}{3}=\frac{2}{3}$
Given: $\mathrm{n}=5$
$\therefore X \sim B\left(5, \frac{1}{3}\right)$
The p.m.f. of $X$ is given by
$P(X=x)={ }^n \mathrm{C}_x p^x q^{n-x}, x=0,1,2,4,5$
i.e. $p(x)={ }^5 C_x\left(\frac{1}{3}\right)^x\left(\frac{2}{3}\right)^{5-x}$
i.e. $p(x)={ }^5 C_x\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{5-x} \quad, x=0,1,2,3,4,5$
$P$ (four or more correct answers) $=P[X \geq 4]$
$ =p(4)+p(5)$
$={ }^5 \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^{5-4}+{ }^5 \mathrm{C}_5\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^{5-5}$
$=5 \times\left(\frac{1}{3}\right)^4 \times\left(\frac{2}{3}\right)^1+1 \times\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^0$
$=\left(\frac{1}{3}\right)^4\left[5 \times \frac{2}{3}+\frac{1}{3}\right]$
$=\left(\frac{1}{3}\right)^4\left[\frac{10}{3}+\frac{1}{3}\right]=\frac{1}{81} \times \frac{11}{3}=\frac{11}{243} $
Hence, the probability of getting four or more correct answers $=\frac{11}{243}$.

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Question 62 Marks

A bag consists of $10$ balls each marked with one of the digits $0$ to $9.$ If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit $0?$

Answer

Let $\mathrm{X}=$ number of balls marked with digit 0 .
$\mathrm{p}=$ probability of drawing a ball from 10 balls marked with the digit 0 .
$\therefore \mathrm{p}=\frac{1}{10}$
and $q=1-p=1-\frac{1}{10}=\frac{9}{10}$
The p.m.f. of $X$ is given by
$ P(X=x)={ }^n \mathrm{C}_x p^x q^{n-x}$
$\text { i.e. } p(x)={ }^4 \mathrm{C}_x\left(\frac{1}{10}\right)^x\left(\frac{9}{10}\right)^{4-x}, x=0,1, \ldots, 4 $
$P($ none of the ball marked with digit 0$)=P(X=0)$
$=p(0)={ }^4 \mathrm{C}_0\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^{4-0}$
$=1 \times 1 \times\left(\frac{9}{10}\right)^2=\left(\frac{9}{10}\right)^4$
Hence, the probability that none of the bulb marked with digit 0 is $\left(\frac{9}{10}\right)^4$

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Maths Solve the Following Question.(2 Marks)

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