Question 15 Marks
A coin is tossed till either a head or $5$ tails are obtained. If a random variable $X$ denotes the necessary number of trials of tossing the coin, then obtain probability distribution of the random variable $X$ and calculate its mean and variance.
AnswerA coin is tossed till either a head or $5$ tails are obtained.
Suppose, $X =$ Number of trials required for tossing a coin.
Here,
$U = {(H), (TH), (TTH), (TTTH), (TTTTH), (TTTTT)}$
$\therefore X = 1, 2, 3, 4, 5, 5$
The following table shows the probability distribution of $X :$
To find the mean and variance of the distribution, the following table is prepared : Mean of the distribution:
$µ = E(X)$
$= \sum x∙ p(x) =\frac{31}{16}$
Variance of the distribution:
$\sigma ^2 = V(X) = E(X^2) – [E(X)]^2$
$= \sum x^2 ∙ p(x) ∙ [\sum x ∙ p(x)]^2$
$= \frac{83}{16}-\left(\frac{31}{16}\right)^{2}$
$=\frac{83}{16}-\frac{961}{256}$
$= \frac{1328-961}{256}$ = $\frac{367}{256}$
Hence, the mean and variance of the distribution obtained are $\frac{31}{16}$ and $\frac{367}{256}$ respectively. View full question & answer→Question 25 Marks
A box contains $4$ red and $2$ blue balls. Three balls are simultaneously drawn at random. If $X$ denotes the number of red bails in the selected balls, find the probability distribution of $X$ and find the expected number of red balls in the selected balls.
AnswerIn a box total $( 4$ red $+2$ blue$) 6$ balls.
$3$ balls are simultaneously drawn at random.
$\therefore$ Total number of primary outcomes $n={ }^6 \mathrm{C}_3=\frac{6 \times 5 \times 4}{3 \times 2 \times 1}=20$
$X=$ No. of red balls in $3$ balls drawn.
There are $4$ red balls in a box.
$\therefore \mathrm{X}=0,1,2,3 .$
The probability distribution of $X$ is obtained as follows :
Expected value of red balls :
$E(X) = Σx ∙ p(x)$
$= 2$
Hence, the expected value of red balls obtained is $2$ balls. View full question & answer→Question 35 Marks
A die is randomly tossed two times. Determine the probability distribution of the sum of the numbers appearing both the times on the die and obtain expected value of the sum.
AnswerA die is randomly tossed two times.
$\therefore $ Total number of outcomes in the sample space U is $n = 6^2 = 36$.
For the sum of the numbers appearing both the times on the die is expressed as $X = (u + v),$ where $(u, v)$ is the element of $U.$
Now, the probability distribution $X$ is obtained as shown in the following table : 
Hence, the expected value of the sum obtained is $7.$
Alternative method :
Expected value of the sum:
$\mathrm{E}(X)=\Sigma x \cdot p(x)=\frac{252}{36}=7$ View full question & answer→Question 45 Marks
The probability distribution of a random variable $X$ is $p (x).$ Variable $X$ can assume the values $x_1 = – 2, x_2 = – 1, x_3 = 1$ and $x_4 = 2$ and if $4p(x _1) = 2p (x_2) = 3p (x_3) = 4p (x_4),$ then obtain mean and variance of this probability distribution.
Answer$p(x)$ is the probability distribution of a random variable $X$.
$x=x_1 ; x_2, x_3, x_4 \text { and } x_1=-2, x_2=-1, x_3=1, x_4=2 .$
$ \sum p(x)=1$
$ \therefore p\left(x_1\right)+p\left(x_2\right)+p\left(x_3\right)+p\left(x_4\right)=1 \ldots \ldots \ldots \ldots \ldots . .(1)$
$ \text { Now, } 4 p\left(x_1\right)=2 p\left(x_2\right)=3 p\left(x_3\right)=4 p\left(x_4\right)$
$ \therefore 4 p\left(x_1\right)=4 p\left(x_4\right), 2 p\left(x_2\right)=4 p\left(x_4\right) \text { and } 3 p\left(x_3\right)=4 p\left(x_4\right)$
$ \therefore p\left(x_1\right)=p\left(x_4\right), p\left(x_2\right)=2 p\left(x_4\right) \text { and }$
$ P\left(x_3\right)=\frac{4 p\left(x_4\right)}{3}$
Putting these values in the result $(1),$
$\mathrm{p}\left(x_4\right)+2 p\left(x_4\right)+\frac{4 p\left(x_4\right)}{3}+p\left(x_4\right)=1$
$ \therefore \frac{3 p\left(x_4\right)+6 p\left(x_4\right)+4 p\left(x_4\right)+3 p\left(x_4\right)}{3}=1$
$ \therefore 16 p\left(x_4\right)=3$
$ \therefore p\left(x_4\right)=\frac{3}{16}$
Now, $P\left(x_1\right)=P\left(x_4\right)=\frac{3}{16}$
$P\left(x_1\right)=2 P\left(x_4\right)$
$ =2 \times \frac{3}{16}$
$ =\frac{6}{16}$
$ P\left(x_3\right)=\frac{4 p\left(x_4\right)}{3}$
$ =\frac{4 \times 3}{3 \times 16}=\frac{4}{16}$
$x_1=-2, x_2=-1, x_3=1$, and $x_4=2$,the probability distribution of random variable $X$ is written as follows :
To find the mean and variance, the table is prepared as follows :
Mean of the distribution :
$\mu=E(X)=\Sigma x \cdot p(x)=-\frac{1}{8}$
Variance of the distribution:
$\sigma^2=V(X)=E\left(X^2\right)-[E(X)]^2$
$ =\sum x^2 \cdot p(x)-\left[\sum x-p(x)\right]^2$
$ =\frac{34}{16}-\left(-\frac{1}{8}\right)^2$
$ =\frac{34}{16}-\frac{1}{64}$
$ =\frac{136-1}{64}=\frac{135}{64}$
Hence, the mean and variance of the distribution obtained are $\frac{1}{8}$ and $\frac{135}{64}$ respectively. View full question & answer→Question 55 Marks
The probability that A obtains the solution of the questions is $60 \%$. Find the probability that he obtains the solution of exact five questions out of six questions.
Answer$\frac{2916}{15625}$
View full question & answer→Question 65 Marks
In a game of chess the probability that $A$ wins the game is $\frac{2}{3}$ Find the probability that out of $4$ trials $A$ win at least in one game.
View full question & answer→Question 75 Marks
In an examination for graduation there are total seven subjects. The probability that a student passes In any of the subject is $0.75$. A student has to pass in all the subject to be a graduate. If $10000$ students appeared In the examination, then find the expected number of graduate student.
View full question & answer→Question 85 Marks
Out of $10$ questions of true or false type. find the probability where at least $8$ questions are true.
View full question & answer→Question 95 Marks
In a lot of $25$ electric bulbs, $5$ bulbs are defective. In a random sample of $5$ bulbs. Find the probability of $(1)$ getting only one defective bulb. $(2)$ getting at least one defective bulb.
Answer$(1) \frac{256}{625},(2) \frac{2101}{3125}$
View full question & answer→Question 105 Marks
The probability that a worker who is working in a chemical factory is affected by a certain diseases is $\frac{1}{3}$. If $5$ workers are selected at random, find the probability that $2$ of them are affected by the disease.
View full question & answer→Question 115 Marks
The probability that a bomb dropped from an aeroplane shoots the target is $\frac{2}{3}$. To destroy a bridge, it should be bombarded $($blasted$)$ with at least $2$ bombs. Find the probability that the bridge is destroyed when $4$ bombs are dropped from an aeroplane on the bridge.
View full question & answer→Question 125 Marks
During monsoon, out of $30$ days it rains on $10$ days. Find the probability that it rains on at least $3$ days within a week.
View full question & answer→Question 135 Marks
The probability that a child has eye defect is $0.25$. There are $192$ families having $3$ children each. In how many families $(1)$ there is not a single child having eye defect? $(2)$ there is at the most one child having eye defect?
Answer$(1)\ 81$ families $(2)\ 162$ families
View full question & answer→Question 145 Marks
The probability that a healthy person lives upto the age of $50$ years is $75 \%$. Out of $200$ healthy persons, $5$ persons are selected at random. Find the probability that $(1)$ none of the selected persons dies up to the age of $50$ years, $(2)$ exactly $2$ of the selected persons die up to the age of $50$ years.
Answer$P =$ probability that a person dies $=\frac{1}{4} (1)\frac{243}{1024}(2) \frac{135}{512}$
View full question & answer→Question 155 Marks
It is assumed that half the population of India is vegetarian. If $100$ enumerators take a sample of $10$ persons each, how many of them will state $($report$)$ that $3$ or less than $3$ persons are vegetarians $?$
Answer$\frac{11}{64}$, number of enumeratirs $=100 \times \frac{11}{64}=17$
View full question & answer→Question 165 Marks
The chance of a person shooting a target is $50 \%$. Find the probability that out of $7$ trials, he shoots the target in at the most $4$ trials. $3$ persons are vegetarians?
View full question & answer→Question 175 Marks
The probability that a worker of a certain factory falls ill is $\frac{1}{3}$ What is the probability that $3$ or more than $3$ workers out of $5$ randomly selected workers of the factory fall ill?
Answer$\frac{\overline{17}}{81}$
View full question & answer→Question 185 Marks
The probability that a student doing job along with studies becomes a graduate is $\frac{2}{5}$. if $5$ such students are selected randomly, find the probability that $(1)$ none of them becomes a graduate, $(2)$ two of them become graduates.
Answer$(1)\ \frac{243}{3125}\ (2)\ \frac{216}{625}$
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