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Maths MCQ

Probability Distributions · Maharashtra Board STD 12 Science (MSBSHSE - English Medium). 101 questions.

Questions · Page 2 of 3

MCQ

MCQ 512 Marks
If three fair coins are tossed, where $X=$ number of heads obtained, then $E(X)$ is
  • A
    $\frac{1}{2}$
  • ✓
    $\frac{3}{2}$
  • C
    $\frac{2}{3}$
  • D
    $\frac{3}{4}$
Answer
Correct option: B.
$\frac{3}{2}$
(B)
$X$ can take values $0,1,2$ and 3 .
$\begin{aligned}& \mathrm{P}(\mathrm{X}=0)=\text { Probability of getting no head }=\frac{1}{8} \\& \mathrm{P}(\mathrm{X}=1)=\text { Probability of getting one head }=\frac{3}{8} \\& \mathrm{P}(\mathrm{X}=2)=\text { Probability of getting two heads }=\frac{3}{8} \\& \mathrm{P}(\mathrm{X}=3)=\text { Probability of getting three heads }=\frac{1}{8}\end{aligned}$
$\therefore \quad \mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot\mathrm{P}\left(x_{\mathrm{i}}\right)$
$\begin{aligned} & =(0)\left(\frac{1}{8}\right)+(1)\left(\frac{3}{8}\right)+(2)\left(\frac{3}{8}\right)+(3)\left(\frac{1}{8}\right) \\ & =0+\frac{3}{8}+\frac{3}{4}+\frac{3}{8}=\frac{12}{8}=\frac{3}{2}\end{aligned}$
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MCQ 522 Marks
The p.m.f. of a r.v. X is given by
X = x0123
P(X = x)$q^3$$3 q^2 p$$3 q p^2$$p^3$
If p+q=1, then E(X)=
  • A
    p
  • B
    2p
  • ✓
    3p
  • D
    4p
Answer
Correct option: C.
3p
(C)
$\mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right)$
$ =0\left(q^3\right)+1\left(3 q^2 p\right)+2\left(3 q p^2\right)+3\left(p^3\right)$
$=3 p q(q+2 p)+3 p^3$
$=3 p q[(p+q)+p]+3 p^3$
$=3 p q(1+p)+3 p^3\quad\ldots[\because p+q=1] $
$=3 p q+3 p^2 q+3 p^3 $
$=3 p q+3 p^2(q+p)$
$=3 p(q+p)\quad\ldots[\because p+q=1]$
$=3 p(1)$
$=3 p$
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MCQ 532 Marks
A random variable X has the following p.m.f.:
X = x012
P(X = x)$\mathrm{q}^2$2pq$\mathrm{p}^2$
If p+q=1, then the variance of X is
  • A
    pq
  • ✓
    2pq
  • C
    $\mathrm{p}^2+\mathrm{q}^2$
  • D
    $\mathrm{p}^2-\mathrm{q}^2$
Answer
Correct option: B.
2pq
(B)
$\mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right)$
$\begin{aligned} & =0\left(q^2\right)+1(2 p q)+2\left(p^2\right) \\ & =2 p q+2 p^2 \\ & =2 p(q+p) \\ & =2 p\quad\ldots[\because p+q=1]\end{aligned}$
$\operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2$
$\begin{aligned} & =0\left(q^2\right)+1^2(2 p q)+2^2\left(p^2\right)-(2 p)^2 \\ & =2 p q+4 p^2-4 p^2 \\ & =2 p q\end{aligned}$
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MCQ 542 Marks
The probability distribution of a discrete random variable X is given by
X-1012
P(X)$\frac{1}{3}$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{3}$
Then, the value of $6 \mathrm{E}\left(\mathrm{X}^2\right)-\operatorname{Var}(\mathrm{X})$ is
  • A
    $\frac{12}{113}$
  • ✓
    $\frac{113}{12}$
  • C
    $\frac{19}{12}$
  • D
    $\frac{1}{2}$
Answer
Correct option: B.
$\frac{113}{12}$
(B)
$ \mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right)=-\frac{1}{3}+0+\frac{1}{6}+\frac{2}{3}=\frac{1}{2}$
$\begin{aligned} & \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ & =\frac{(-1)^2}{3}+0+\frac{1^2}{6}+\frac{2^2}{3}-\left(\frac{1}{2}\right)^2 \\ & =\frac{1}{3}+\frac{1}{6}+\frac{4}{3}-\frac{1}{4} \\ & =\frac{11}{6}-\frac{1}{4}=\frac{19}{12} \\  \therefore & 6 \mathrm{E}\left(\mathrm{X}^2\right)-\operatorname{Var}(\mathrm{X}) \\ & =6\left(\frac{11}{6}\right)-\frac{19}{12} \\ & =11-\frac{19}{12} \\ & =\frac{113}{12}\end{aligned}$
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MCQ 552 Marks
The probability distribution of a random variable X is given below.
X = k01234
P(X = k)0.10.40.30.20
The variance of X is
  • A
    1.6
  • B
    0.24
  • ✓
    0.84
  • D
    0.75
Answer
Correct option: C.
0.84
(C)
$\mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right)$
$\begin{aligned}&=0(0.1)+1(0.4)+2(0.3)+3(0.2)+4(0) \\ &=0+0.4+0.6+0.6+0 \\ &=1.6 \\ & \text { Variance }=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ &= 0^2(0.1)+1^2(0.4)+2^2(0.3)+3^2(0.2)+4^2(0)-1.6^2 \\ &= 0+0.4+1.2+1.8-2.56 \\ &= 0.84\end{aligned}$
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MCQ 562 Marks
The probability distribution of a random variable X is given below. If its mean is 4.2 , then find the values of a and b .
X = x123456
P(X = x)aaabb0.3
  • A
    $0.3,0.2$
  • B
    $0.1,0.4$
  • ✓
    $0.1,0.2$
  • D
    $0.2,0.1$
Answer
Correct option: C.
$0.1,0.2$
(C)
Since $\sum_{x=1}^6 \mathrm{P}(\mathrm{X}=x)=1$,
$\begin{aligned}& a+a+a+b+b+0.3=1 \\& \Rightarrow 3 a+2 b=0.7\quad\ldots(i) \\& \text { Mean }=a+2 a+3 a+4 b+5 b+6(0.3) \\& \Rightarrow 4.2=6 a+9 b+1.8 \\& \Rightarrow 6 a+9 b=2.4\quad\ldots(ii)\end{aligned}$
On solving (i) and (ii), we get
$a=0.1, b=0.2$
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MCQ 572 Marks
Two dice are rolled. If a random variable X is defined as the absolute difference of the two numbers that appear on them, then the mean of X is
  • A
    $0$
  • B
    $\frac{13}{18}$
  • C
    $\frac{19}{9}$
  • ✓
    $\frac{35}{18}$
Answer
Correct option: D.
$\frac{35}{18}$
(D)
Possible values of X are $0,1,2,3,4,5$
$\therefore$ Probability distribution of X is given as
X = x012345
P(X = x)$\frac{6}{36}$$\frac{10}{36}$$\frac{8}{36}$$\frac{6}{36}$$\frac{4}{36}$$\frac{2}{36}$
$\begin{aligned} \therefore \text { Mean } & =0+1\left(\frac{10}{36}\right)+2\left(\frac{8}{36}\right)+3\left(\frac{6}{36}\right)+4\left(\frac{4}{36}\right)+5\left(\frac{2}{36}\right) \\ & =\frac{35}{18}\end{aligned}$
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MCQ 582 Marks
If the probability mass function of a discre random variable X is
$\begin{aligned}\mathrm{P}(x) & =\frac{\mathrm{C}}{x^3} ; \quad x=1,2,3 \\& =0 ; \quad \text { otherwise }\end{aligned}$
Then $\mathrm{E}(\mathrm{X})=$
  • A
    $\frac{343}{297}$
  • ✓
    $\frac{294}{251}$
  • C
    $\frac{297}{294}$
  • D
    $\frac{251}{294}$
Answer
Correct option: B.
$\frac{294}{251}$
(B)
$\mathrm{P}(1)=\frac{\mathrm{C}}{1^3}, \mathrm{P}(2)=\frac{\mathrm{C}}{2^3}, \mathrm{P}(3)=\frac{\mathrm{C}}{3^3}$
$\text { Since, } \mathrm{P}(1)+\mathrm{P}(2)+\mathrm{P}(3)=1$
$\therefore \frac{\mathrm{C}}{1^3}+\frac{\mathrm{C}}{2^3}+\frac{\mathrm{C}}{3^3}=1$
$\Rightarrow \mathrm{C}\left(\frac{1}{1}+\frac{1}{8}+\frac{1}{27}\right)=1$
$\Rightarrow \mathrm{C}\left(\frac{216+27+8}{216}\right)=1$
$\Rightarrow \mathrm{C}=\frac{216}{251}$
$\therefore E ( X )=\sum x_{ i } \cdot P \left(x_{ i }\right)$
$=(1) \frac{\mathrm{C}}{1^3}+(2) \frac{\mathrm{C}}{2^3}+(3) \frac{\mathrm{C}}{3^3}$
$=\mathrm{C}\left(1+\frac{1}{4}+\frac{1}{9}\right)=\mathrm{C}\left(\frac{36+9+4}{36}\right) $
$=\frac{216}{251} \times \frac{49}{36}$
$=\frac{294}{251}$
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MCQ 592 Marks
The p.m.f. of a r.v. X is
$P(x)=\left\{\begin{array}{cc}k x^2 ; & x=1,2,3,4 \\
0 ; & \text { otherwise }\end{array}\right.$
Then, $E(X)=$
  • A
    $\frac{7}{3}$
  • B
    $\frac{5}{3}$
  • ✓
    $\frac{10}{3}$
  • D
    $\frac{8}{3}$
Answer
Correct option: C.
$\frac{10}{3}$
(C)
X = x1234
P(X = x)k4k9k16k
$\text {Since,} \mathrm{P}(1)+\mathrm{P}(2)+\mathrm{P}(3)+\mathrm{P}(4)=1$
$\begin{aligned} & \therefore \mathrm{k}+4 \mathrm{k}+9 \mathrm{k}+16 \mathrm{k}=1 \\ & \Rightarrow 30 \mathrm{k}=1 \\ & \Rightarrow \mathrm{k}=\frac{1}{30} \\ & \therefore \mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right) \\ & =1 \cdot \frac{1}{30}+2 \cdot \frac{4}{30}+3 \cdot \frac{9}{30}+4 \cdot \frac{16}{30} \\ & =\frac{100}{30}=\frac{10}{3}\end{aligned}$
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MCQ 602 Marks
The p.m.f. of a random variable X is
$\begin{aligned}\mathrm{P}(x) & =\frac{3-x}{10}, & & x=-1,0,1,2 \\& =0, & & \text { otherwise }\end{aligned}$
The value of $\mathrm{E}(\mathrm{X})$ is
  • ✓
    $0$
  • B
    1
  • C
    2
  • D
    3
Answer
Correct option: A.
$0$
(A)
The p.m.f. of the r.v. X is as follows:
X = x-1012
P(X = x)$\frac{2}{5}$$\frac{3}{10}$$\frac{1}{5}$$\frac{1}{10}$
$\begin{aligned} \therefore \quad \mathrm{E}(\mathrm{X}) & =\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right) \\ & =-1\left(\frac{2}{5}\right)+0+1\left(\frac{1}{5}\right)+2\left(\frac{1}{10}\right)=0\end{aligned}$
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MCQ 612 Marks
A random variable X has the following probability distribution:
$\mathrm{X}=x_i$1234
P($\mathrm{X}=x_i$)0.10.20.30.4
The mean and the standard deviation are respectively
  • A
    3 and 2
  • ✓
    3 and 1
  • C
    3 and $\sqrt{3}$
  • D
    2 and 1
Answer
Correct option: B.
3 and 1
(B)
Mean $=\mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right)$
$=1(0.1)+2(0.2)+3(0.3)+4(0.4)=3$
$\begin{aligned}& \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\& =1^2(0.1)+2^2(0.2)+3^2(0.3)+4^2(0.4)-(3)^2 \\& =0.1+0.8+2.7+6.4-9=10-9=1\end{aligned}$
$\therefore \quad$ S.D. $=1$
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MCQ 622 Marks
If X is a random variable with probability distribution as given below:
X0123
P(X)k3k3kk
Then, the variance of X is
  • A
    $\frac{1}{8}$
  • B
    $\frac{23}{7}$
  • C
    $\frac{24}{27}$
  • ✓
    $\frac{3}{4}$
Answer
Correct option: D.
$\frac{3}{4}$
(D)
The sum of all the probabilities in a probability distribution is always unity.
$ \therefore \quad \mathrm{k}+3 \mathrm{k}+3 \mathrm{k}+\mathrm{k}=1$
$\begin{aligned} & \Rightarrow 8 \mathrm{k}=1 \\& \Rightarrow \mathrm{k}=\frac{1}{8} \\& \mathrm{E}(\mathrm{X})-\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right) \\& =0\left(\frac{1}{8}\right)+1\left(\frac{3}{8}\right)+2\left(\frac{3}{8}\right)+3\left(\frac{1}{8}\right)=\frac{3}{2} \\& \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\& =0^2\left(\frac{1}{8}\right)+1^2\left(\frac{3}{8}\right)+2^2\left(\frac{3}{8}\right)+3^2\left(\frac{1}{8}\right)-\left(\frac{3}{2}\right)^2 \\& =\frac{3}{4}\end{aligned}$
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MCQ 632 Marks
A random variable X has the following probability distribution:
$X = x$-2-10123
P$(X = x$)0.1k0.22k0.3k
Then, the expected value of X is
  • A
    0.6
  • B
    0.5
  • C
    0.7
  • ✓
    0.8
Answer
Correct option: D.
0.8
(D)
The sum of all the probabilities in a probability distribution is always unity.
$\begin{array}{ll}\therefore 0.1+\mathrm{k}+0.2+2 \mathrm{k}+0.3+\mathrm{k}=1 \\ \Rightarrow 0.6+4 \mathrm{k}=1 \\ \Rightarrow 4 \mathrm{k}=0.4 \\ \Rightarrow \mathrm{k}=0.1 \\\therefore \mathrm{E}(\mathrm{X})=\sum x_i \cdot \mathrm{P}\left(x_i\right) \\ =(-2)(0.1)+(-1)(0.1)+0(0.2)+1(2 \times 0.1)+2(0.3)+3(0.1)=0.8\end{array}$
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MCQ 642 Marks
A random variable $X$ has the following probability distribution:
$X = x$1234
P$(X = x)$0.20.10.3k
Then, the variance of X=
  • ✓
    1.29
  • B
    1.31
  • C
    1.27
  • D
    1.23
Answer
Correct option: A.
1.29
(A)
The sum of all the probabilities in a probability distribution is always unity.
$\begin{array}{ll}\therefore & 0.2+0.1+0.3+\mathrm{k}=1 \\\therefore & \mathrm{k}=1-0.6=0.4 \\& \mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right)\end{array}$
$\begin{aligned} & =1(0.2)+2(0.1)+3(0.3)+4(0.4) \\ & =0.2+0.2+0.9+1.6=2.9\end{aligned}$
$\begin{aligned} \operatorname{Var}(\mathrm{X}) & =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ & =(1)^2(0.2)+(2)^2(0.1)+(3)^2(0.3)+(4)^2(0.4)-(2.9)^2 \\ & =0.2+0.4+2.7+6.4-8.41 \\ & =9.7-8.41=1.29\end{aligned}$
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MCQ 652 Marks
The c.d.f. of a discrete r.v. X is
X = x-3-1013579
F(x)0.10.30.50.650.750.850.901
Then $\mathrm{P}(\mathrm{X} \leq 3 \mid \mathrm{X}>0)=$
  • ✓
    $\frac{1}{2}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{1}{4}$
  • D
    $\frac{1}{5}$
Answer
Correct option: A.
$\frac{1}{2}$
(A)
$P(X=1)=F(1)-F(0)=0.65-0.5=0.15$
$P(X=3)=F(3)-F(1)=0.75-0.65=0.10$
$P(X=5)=0.85-0.75=0.10$
$P(X=7)=0.90-0.85=0.05$
$P(X=9)=1-0.90=0.10$
$\therefore P(X \leq 3 \mid X>0)$
$ =\frac{P(X=1)+P(X=3)}{P(X=1)+P(X=3)+P(X=5)+P(X=7)+P(X=9)}$
$ =\frac{0.15+0.1}{0.15+0.1+0.1+0.05+0.1}$
$=\frac{0.25}{0.50}=\frac{1}{2}$
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MCQ 662 Marks
The c.d.f. of a discretc r.v. $x$ is
X123456
F$(x)$0.180.430.540.680.891.00
Then P (1 < x < 5) =
  • A
    0.25
  • ✓
    0.5
  • C
    0.75
  • D
    0.36
Answer
Correct option: B.
0.5
(B)
$ \mathrm{P}(x=2)=\mathrm{F}(2)-\mathrm{F}(1)=0.43-0.18=0.25$
$\mathrm{P}(x-3)-\mathrm{F}(3)-\mathrm{F}(2)-0.54-0.43-0.11$
$\mathrm{P}(x=4)=\mathrm{F}(4)-\mathrm{F}(3)=0.68-0.54=0.14$
$\therefore P (1 < x<5)= P (x=2)+ P (x=3)+ P (x=4)$
$=0.25+0.11+0.14$
$=0.50$
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MCQ 672 Marks
Five defective mangoes are accidently mixed with 15 good ones. Four mangoes are drawn at random from this lot. Then the probability distribution of the number of defective mangoes is
  • A
    X01234
    P(X)$\frac{85}{323}$$\frac{5}{323}$$\frac{1}{969}$$\frac{2}{969}$$\frac{3}{969}$
  • B
    X01234
    P(X)$\frac{91}{323}$$\frac{85}{969}$$\frac{3}{323}$$\frac{1}{969}$$\frac{3}{969}$
  • ✓
    X01234
    P(X)$\frac{91}{323}$$\frac{455}{969}$$\frac{70}{323}$$\frac{10}{323}$$\frac{1}{969}$
  • D
    X01234
    P(X)$\frac{455}{969}$$\frac{85}{323}$$\frac{263}{323}$$\frac{25}{969}$$\frac{2}{969}$
Answer
Correct option: C.
X01234
P(X)$\frac{91}{323}$$\frac{455}{969}$$\frac{70}{323}$$\frac{10}{323}$$\frac{1}{969}$
(C)
Let X denote the number of defective mangoes from the bag. $X$ can take values 0,1 , 2,3 and 4 .
$\mathrm{P}(\mathrm{X}=0)=$ Probability of getting no defective mango $=\frac{{ }^{15} \mathrm{C}_4}{{ }^{20} \mathrm{C}_4}=\frac{91}{323}$
$\mathrm{P}(\mathrm{X}=1)=$ Probability of getting one defective mango $=\frac{{ }^5 \mathrm{C}_1 \times{ }^{15} \mathrm{C}_3}{{ }^{20} \mathrm{C}_4}=\frac{455}{969}$
$\mathrm{P}(\mathrm{X}=2)=$ Probability of getting two defective mangoes $=\frac{{ }^5 \mathrm{C}_2 \times{ }^{15} \mathrm{C}_2}{{ }^{20} \mathrm{C}_4}=\frac{70}{323}$
$\mathrm{P}(\mathrm{X}=3)=$ Probability of getting three defective mangoes $=\frac{{ }^5 \mathrm{C}_3 \times{ }^{15} \mathrm{C}_1}{{ }^{20} \mathrm{C}_4}=\frac{10}{323}$
$\mathrm{P}(\mathrm{X}=4)=$ Probability of getting four defective mangoes $=\frac{{ }^5 \mathrm{C}_4}{{ }^{20} \mathrm{C}_4}=\frac{1}{969}$
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MCQ 682 Marks
A bag contains 4 red and 6 black balls. Three balls are drawn at random. Then the probability distribution of the number of red balls is
  • A
    X0123
    P(X)$\frac{1}{9}$$\frac{2}{9}$$\frac{5}{6}$$\frac{4}{6}$
  • ✓
    X0123
    P(X)$\frac{1}{6}$$\frac{1}{2}$$\frac{3}{10}$$\frac{1}{30}$
  • C
    X0123
    P(X)$\frac{1}{5}$$\frac{2}{3}$$\frac{5}{9}$$\frac{6}{9}$
  • D
    X0123
    P(X)$\frac{1}{6}$$\frac{2}{3}$$\frac{4}{10}$$\frac{3}{10}$
Answer
Correct option: B.
X0123
P(X)$\frac{1}{6}$$\frac{1}{2}$$\frac{3}{10}$$\frac{1}{30}$
(B)
Let X denote the number of red balls drawn from the bag. There are 4 red balls and $X$ can take values $0,1,2$ and 3 .
$\mathrm{P}(\mathrm{X}=0)=$ Probability of getting no red ball
$=\frac{{ }^6 \mathrm{C}_3}{{ }^{10} \mathrm{C}_3}=\frac{1}{6}$
$\mathrm{P}(\mathrm{X}=1)=$ Probability of getting one red ball
$=\frac{{ }^4 \mathrm{C}_1 \times{ }^6 \mathrm{C}_2}{{ }^{10} \mathrm{C}_3}=\frac{1}{2}$
$\mathrm{P}(\mathrm{X}=2)=$ Probability of getting two red balls
$=\frac{{ }^4 \mathrm{C}_2 \times{ }^6 \mathrm{C}_1}{{ }^{10} \mathrm{C}_3}=\frac{3}{10}$
$\mathrm{P}(\mathrm{X}=3)=$ Probability of getting three red balls
$=\frac{{ }^4 \mathrm{C}_3}{{ }^{10} \mathrm{C}_3}=\frac{1}{30}$
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MCQ 692 Marks
Three coins are tossed. Then the probability distribution of number of heads is
  • A
    X 0 1 2 3
    P(X) $\frac{1}{8}$ $\frac{5}{8}$ $\frac{5}{8}$ $\frac{1}{8}$
  • B
    X0123
    P(X)$\frac{2}{8}$$\frac{1}{8}$$\frac{1}{8}$$\frac{2}{8}$
  • C
    X0123
    P(X)$\frac{1}{5}$$\frac{2}{5}$$\frac{2}{5}$$\frac{1}{5}$
  • ✓
    X0123
    P(X)$\frac{1}{8}$$\frac{3}{8}$$\frac{3}{8}$$\frac{1}{8}$
Answer
Correct option: D.
X0123
P(X)$\frac{1}{8}$$\frac{3}{8}$$\frac{3}{8}$$\frac{1}{8}$
(D)
Let X denotes the number of heads.
Thus, the possible values of X are $0,1,2$ and 3 .
$\begin{aligned}\mathrm{P}(\mathrm{X}=0) & =\mathrm{P}(\text { getting no head }) \\& =\mathrm{P}(\mathrm{TTT})=\frac{1}{8} \\\mathrm{P}(\mathrm{X}=1) & =\mathrm{P}(\text { getting one head }) \\& =\mathrm{P}(\mathrm{HTT}, \text { THT, TTH })=\frac{3}{8} \\\mathrm{P}(\mathrm{X}=2) & =\mathrm{P}(\text { getting two heads }) \\& =\mathrm{P}(\mathrm{HHT}, \text { THH, HTH })=\frac{3}{8} \\\mathrm{P}(\mathrm{X}=3) & =\mathrm{P}(\text { getting three heads }) \\& =\mathrm{P}(\mathrm{HHH})=\frac{1}{8}\end{aligned}$
$\therefore$ Option (D) is the correct answer.
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MCQ 702 Marks
If the probability function of a random variable $X$ is defined by $P(X=k)=a\left(\frac{k+1}{2^k}\right)$ for $k=0,1,2,3,4,5$, then the probability that X takes a prime value is
  • A
    $\frac{13}{20}$
  • ✓
    $\frac{23}{60}$
  • C
    $\frac{11}{20}$
  • D
    $\frac{19}{60}$
Answer
Correct option: B.
$\frac{23}{60}$
(B)
X = k012345
P(X = k)aa$\frac{3a}{4}$$\frac{4a}{8}$$\frac{5a}{16}$$\frac{6a}{32}$
$\begin{aligned}& \text { Since } \sum_{\mathrm{k}=0}^5 \mathrm{P}(\mathrm{X}=\mathrm{k})=1, \\& \mathrm{a}+\mathrm{a}+\frac{3 \mathrm{a}}{4}+\frac{4 \mathrm{a}}{8}+\frac{5 \mathrm{a}}{16}+\frac{6 \mathrm{a}}{32}=1 \\& \Rightarrow \frac{15}{4} \mathrm{a}=1 \Rightarrow \mathrm{a}=\frac{4}{15}\end{aligned}$
Now, $\mathrm{P}(\mathrm{X}=$ prime value $)$
$\begin{aligned}& =P(X=2)+P(X=3)+P(X=5) \\& =\frac{3 a}{4}+\frac{4 a}{8}+\frac{6 a}{32} \\& =\frac{23 a}{16} \\& =\frac{23}{16} \times \frac{4}{15}=\frac{23}{60}\end{aligned}$
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MCQ 712 Marks
Let $X$ be a random variable which assumes values $\quad x_1, x_2, x_3, x_4$ such that $2 \mathrm{P}\left(\mathrm{X}=x_1\right)=3 \mathrm{P}\left(\mathrm{X}=x_2\right)=\mathrm{P}\left(\mathrm{X}=x_3\right)=5 \mathrm{P}\left(\mathrm{X}=x_4\right)$, Then, the probability distribution of X is
  • ✓
    X$x_1$$x_2$$x_3$$x_4$
    P(X)$\frac{15}{61}$$\frac{10}{61}$$\frac{30}{61}$$\frac{6}{61}$
  • B
    X$x_1$$x_2$$x_3$$x_4$
    P(X)$\frac{5}{16}$$\frac{4}{16}$$\frac{2}{16}$$\frac{6}{16}$
  • C
    X$x_1$$x_2$$x_3$$x_4$
    P(X)$\frac{3}{14}$$\frac{4}{14}$$\frac{7}{14}$$\frac{1}{14}$
  • D
    X$x_1$$x_2$$x_3$$x_4$
    P(X)$\frac{10}{31}$$\frac{15}{31}$$\frac{5}{31}$$\frac{2}{31}$
Answer
Correct option: A.
X$x_1$$x_2$$x_3$$x_4$
P(X)$\frac{15}{61}$$\frac{10}{61}$$\frac{30}{61}$$\frac{6}{61}$
(A)
Let $\mathrm{P}\left(\mathrm{X}=x_3\right)=\mathrm{k}$. Then $\mathrm{P}\left(\mathrm{X}=x_1\right)=\frac{\mathrm{k}}{2}$,$\mathrm{P}\left(\mathrm{X}=x_2\right)=\frac{\mathrm{k}}{3}, \mathrm{P}\left(\mathrm{X}=x_4\right)=\frac{\mathrm{k}}{5}$
Since,
$\begin{aligned} \mathrm{P}\left(\mathrm{X}=x_1\right)+\mathrm{P}\left(\mathrm{X}=x_2\right)+\mathrm{P} & \left(\mathrm{X}=x_3\right)+\mathrm{P}\left(\mathrm{X}=x_4\right)=1\end{aligned}$
$\begin{aligned}&\therefore \quad \frac{\mathrm{k}}{2}+\frac{\mathrm{k}}{3}+\mathrm{k}+\frac{\mathrm{k}}{5}=1 \quad \Rightarrow \mathrm{k}=\frac{30}{61}\\&\therefore \quad \text { option }(A) \text { is the correct answer. }\end{aligned}$
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MCQ 722 Marks
Let X denote the number of hours that a bus travels in a city during a randomly selected day. The probability that X can take the value $x$ has the following form, where $k$ is some unknown constant.
$\mathrm{P}(\mathrm{X}=x)= \begin{cases}0.1, & \text { if } x=0 \\ \mathrm{kx}, & \text { if } x=1 \text { or } 2 \\ \mathrm{k}(5-x), & \text { if } x=3 \text { or } 4 \\ 0, & \text { otherwise }\end{cases}$
Then the value of $k$ is
  • A
    0.35
  • B
    0.3
  • ✓
    0.15
  • D
    0.2
Answer
Correct option: C.
0.15
(C)
The probability distribution of X is
X01234
P(X)0.1k2k2kk
$\begin{aligned} & \text {Since} \sum_{x=0}^4 \mathrm{P}(\mathrm{X}=x)=1 \\ & 0.1+\mathrm{k}+2 \mathrm{k}+2 \mathrm{k}+\mathrm{k}=1 \\ & \Rightarrow 6 \mathrm{k}=0.9 \quad \Rightarrow \mathrm{k}=0.15\end{aligned}$
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MCQ 732 Marks
The p.m.f. of a random variable $X$ is
X = x12345
P(X = x)$\frac{1}{20}$$\frac{3}{20}$ab$\frac{1}{20}$
If $b=2 a$, then
  • A
    $\mathrm{a}=\frac{1}{2}, b=\frac{1}{3}$
  • B
    $\mathrm{a}=\frac{1}{3}, \mathrm{~b}=\frac{1}{2}$
  • ✓
    $\mathrm{a}=\frac{1}{4}, \mathrm{~b}=\frac{1}{2}$
  • D
    $\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{1}{4}$
Answer
Correct option: C.
$\mathrm{a}=\frac{1}{4}, \mathrm{~b}=\frac{1}{2}$
(C)
$ \text { Since } \sum_{x=1}^5 \mathrm{P}(\mathrm{X}=x)=1$
$\begin{aligned} & \frac{1}{20}+\frac{3}{20}+\mathrm{a}+\mathrm{b}+\frac{1}{20}=1 \\ & \Rightarrow \mathrm{a}+\mathrm{b}=1-\frac{5}{20} \\ & \Rightarrow \mathrm{a}+2 \mathrm{a}=1-\frac{1}{4} \quad \ldots[\because \mathrm{~b}=2 \mathrm{a} \text { (given) }] \\ & \Rightarrow 3 \mathrm{a}=\frac{3}{4} \\ & \Rightarrow \mathrm{a}=\frac{1}{4} \text { and } \mathrm{b}=2\left(\frac{1}{4}\right)=\frac{1}{2}\end{aligned}$
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MCQ 742 Marks
The p.m.f. of a r.v. $X$ is as follows:
$\mathrm{P}(X=0)=3 k^3, P(X=1)=4 k-10 k^2$
$\mathrm{P}(\mathrm{X}=2)=5 \mathrm{k}-1, \mathrm{P}(\mathrm{X}=x)=0$ for any other value of x. Then the value of k is
  • ✓
    $\frac{1}{3}$
  • B
    $\frac{2}{3}$
  • C
    1
  • D
    2
Answer
Correct option: A.
$\frac{1}{3}$
(A)
$ \text {Since, } \sum_{x=0}^2 \mathrm{P}(\mathrm{X}=x)=1$
$\begin{aligned}& \therefore 3 \mathrm{k}^3+4 \mathrm{k}-10 \mathrm{k}^2+5 \mathrm{k}-1=1 \\& \Rightarrow 3 \mathrm{k}^3-10 \mathrm{k}^2+9 \mathrm{k}-2=0 \\& \Rightarrow(\mathrm{k}-1)(\mathrm{k}-2)(3 \mathrm{k}-1)=0 \\& \Rightarrow \mathrm{k}=1 \text { or } \mathrm{k}=2 \text { or } \mathrm{k}=\frac{1}{3}\end{aligned}$
For $\mathrm{k}=1$ or $\mathrm{k}=2, \mathrm{P}(\mathrm{X}=1)<0$, which is not possible.
$\therefore \quad \mathrm{k}=\frac{1}{3}$
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MCQ 752 Marks
A random variable ' X ' has the following probability distribution
X1234567
P(X)k -13kk3k$\mathrm{3k}^2$$\mathrm{k}^2$$\mathrm{k}^2+k$
Then the value of k is
  • A
    -2
  • B
    $\frac{1}{10}$
  • ✓
    $\frac{1}{5}$
  • D
    $\frac{2}{7}$
Answer
Correct option: C.
$\frac{1}{5}$
(C)
Since $\sum_{x=1}^7 \mathrm{P}(\mathrm{X}=x)=1$,
$\begin{aligned}& \mathrm{k}-1+3 \mathrm{k}+\mathrm{k}+3 \mathrm{k}+3\mathrm{k}^2+\mathrm{k}^2+\mathrm{k}^2+\mathrm{k}=1 \\& \Rightarrow 5 \mathrm{k}^2+9 \mathrm{k}-2=0 \\& \Rightarrow 5 \mathrm{k}^2+10 \mathrm{k}-\mathrm{k}-2=0 \\& \Rightarrow 5 \mathrm{k}(\mathrm{k}+2)-1(\mathrm{k}+2)=0 \\& \Rightarrow(5 \mathrm{k}-1)(\mathrm{k}+2)=0 \\& \Rightarrow \mathrm{k}=\frac{1}{5} \quad \ldots[\because \mathrm{k}=-2 \text { is not possible }]\end{aligned}$
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MCQ 762 Marks
A random variable $X$ has the following probability distribution:
X01234567
P(X)0P2P2P3P$\mathrm{P}^2$$2 \mathrm{P}^2$$7 \mathrm{P}^2+\mathrm{P}$

The value of P is
  • A
    -1
  • ✓
    $\frac{1}{10}$
  • C
    $-\frac{1}{10}$
  • D
    1
Answer
Correct option: B.
$\frac{1}{10}$
(B)
$ \text {Since, } \sum_{x=0}^7 \mathrm{P}(\mathrm{X}=x)=1$
$\begin{aligned} & \therefore 0+\mathrm{P}+2 \mathrm{P}+2 \mathrm{P}+3 \mathrm{P}+\mathrm{P}^2+2 \mathrm{P}^2+7 \mathrm{P}^2+\mathrm{P}=1 \\ & \therefore 10 \mathrm{P}^2+9 \mathrm{P}-1=0 \\ & \therefore(\mathrm{P}+1)(10 \mathrm{P}-1)=0 \\ & \therefore \mathrm{P}=\frac{1}{10} \ldots[\because \mathrm{P} \geq 0, \mathrm{P}+1 \neq 0]\end{aligned}$
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MCQ 772 Marks
$\mathrm{P}(x)=\left\{\begin{array}{cl}\mathrm{k}\binom{4}{x}, & x=0,1,2,3,4 ; \mathrm{k}>0 \\0, & \text { otherwise }\end{array}\right.$
is p.m.f. of a r.v. X. Then $k$ is equal to
  • A
    $\frac{1}{2}$
  • B
    $\frac{1}{4}$
  • C
    $\frac{1}{8}$
  • ✓
    $\frac{1}{16}$
Answer
Correct option: D.
$\frac{1}{16}$
(D)
X = x01234
P$(X = x)$k4k6k4kk

$\text {Since} \sum_{x=0}^4 \mathrm{P}(\mathrm{X}=x)=1$
$\begin{aligned} & \mathrm{k}+4 \mathrm{k}+6 \mathrm{k}+4 \mathrm{k}+\mathrm{k}=1 \\ & \Rightarrow 16 \mathrm{k}=1 \\ & \Rightarrow \mathrm{k}=\frac{1}{16}\end{aligned}$
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MCQ 782 Marks
A random variable X has the following probability distribution:
X = x-3-2-10123
P$(X = x)$0.050.100.150.200.250.150.10
Then $\mathrm{P}(\mathrm{X}$ is odd $)=$
  • A
    0.45
  • ✓
    0.55
  • C
    0.65
  • D
    0.75
Answer
Correct option: B.
0.55
(B)
$\begin{aligned} & \mathrm{P}(\mathrm{X} \text { is odd }) \\ & =\mathrm{P}(\mathrm{X}=-3)+\mathrm{P}(\mathrm{X}=-1)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=3) \\ & =0.05+0.15+0.25+0.10=0.55\end{aligned}$
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MCQ 792 Marks
A random variable X has the following probability distribution:
$X = x$0123456
$P(X = x)$k3k5k7k9k11k13k
Then, the value of $\mathrm{P}(0<\mathrm{X}<4)$ is
  • A
    $\frac{11}{49}$
  • ✓
    $\frac{15}{49}$
  • C
    $\frac{20}{49}$
  • D
    $\frac{40}{49}$
Answer
Correct option: B.
$\frac{15}{49}$
(B)
$\text {Since} \sum_{x=0}^6 \mathrm{P}(\mathrm{X}=x)=1,$
$\begin{aligned} & \mathrm{k}+3 \mathrm{k}+5 \mathrm{k}+7 \mathrm{k}+9 \mathrm{k}+11 \mathrm{k}+13 \mathrm{k}=1 \\ & \Rightarrow 49 \mathrm{k}=1 \Rightarrow \mathrm{k}=\frac{1}{49} \\ & \therefore \mathrm{P}(0<\mathrm{X}<4) \\ & =\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3) \\ &=3 \mathrm{k}+5 \mathrm{k}+7 \mathrm{k}=15 \mathrm{k}=\frac{15}{49}\end{aligned}$
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MCQ 802 Marks
A random variable $X$ has the following probability distribution:
$X = x$01234
P$(X = x)$k3k5k2kk
Then the value of $\mathrm{P}(\mathrm{X} \geq 2)$ is
  • A
    $\frac{1}{3}$
  • ✓
    $\frac{2}{3}$
  • C
    $\frac{3}{4}$
  • D
    $\frac{1}{4}$
Answer
Correct option: B.
$\frac{2}{3}$
(B)
Since, $\sum_{x=0}^4 \mathrm{P}(\mathrm{X}=x)=1$,
$\begin{aligned}& k+3 k+5 k+2 k+k=1 \\& \Rightarrow 12 k=1 \quad \Rightarrow k=\frac{1}{12}\end{aligned}$
Now, $P(X \geq 2)=P(X=2)+P(X=3)+P(X=4)$
$\begin{aligned} & =5 \mathrm{k}+2 \mathrm{k}+\mathrm{k} \\ & =8 \mathrm{k}=8\left(\frac{1}{12}\right)=\frac{2}{3}\end{aligned}$
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MCQ 812 Marks
A random variable X has the following probability distribution:
$X = x$1234
P$(X = x)$k2k3k4k
The value of $\mathrm{P}(\mathrm{X}<3)$ is
  • A
    0.1
  • B
    0.2
  • ✓
    0.3
  • D
    0.4
Answer
Correct option: C.
0.3
(C)
Since $\sum_{x=1}^4 \mathrm{P}(\mathrm{X}=x)=1$,
$\begin{aligned}& \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}+4 \mathrm{k}=1 \\& \Rightarrow 10 \mathrm{k}=1 \\& \Rightarrow \mathrm{k}=\frac{1}{10}\end{aligned}$
Now, $\mathrm{P}(\mathrm{X}<3)=\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)$
$\begin{aligned} & =\mathrm{k}+2 \mathrm{k} \\ & =3 \mathrm{k}=3\left(\frac{1}{10}\right)=0.3\end{aligned}$
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MCQ 822 Marks
The p.d.f. of a continuous random variable $X$ is
$\begin{aligned}\mathrm{f}(x) & =\frac{x^2}{3}, & & -1<x<2 \\& =0, & & \text { otherwise }\end{aligned}$
Then the c.d.f. of X is given by
  • A
    $\mathrm{F}(x)=x^3+\frac{1}{9}$
  • B
    $\mathrm{F}(x)=\frac{x^3}{3}+\frac{1}{9}$
  • ✓
    $\mathrm{F}(x)=\frac{x^3}{9}+\frac{1}{9}$
  • D
    $\mathrm{F}(x)=\frac{x^3}{9}-\frac{1}{9}$
Answer
Correct option: C.
$\mathrm{F}(x)=\frac{x^3}{9}+\frac{1}{9}$
(C)
The c.d.f. of X is
$\mathrm{F}(x)=\int_{-1}^x \frac{x^2}{3} \mathrm{~d} x=\frac{1}{3}\left[\frac{x^3}{3}\right]_{-1}^x$
$=\frac{1}{3}\left(\frac{x^3}{3}+\frac{1}{3}\right)=\frac{x^3}{9}+\frac{1}{9}$
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MCQ 832 Marks
The p.d.f. of a continuous random variable $X$ is
$ \begin{aligned} \mathrm{f}(x) & =\frac{x}{8}, 0<x<4 \\ & =0, \quad \text { otherwise } \end{aligned} $
Then the value of $\mathrm{P}(\mathrm{X}>3)$ is
  • A
    $\frac{3}{16}$
  • B
    $\frac{5}{16}$
  • ✓
    $\frac{7}{16}$
  • D
    $\frac{9}{16}$
Answer
Correct option: C.
$\frac{7}{16}$
(C)
$\mathrm{P}(\mathrm{X}>3)=\int_3^4 \frac{x}{8} \mathrm{~d} x=\frac{1}{8}\left[\frac{x^2}{2}\right]_3^4=\frac{7}{16}$
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MCQ 842 Marks
If the p.d.f. of a random variable X is
$\mathrm{f}(x)=\left\{\begin{array}{cc}0.5 x, & 0 \leq x \leq 2 \\ 0, & \text { otherwise }\end{array}\right.$
then the value of $\mathrm{P}(0.5 \leq \mathrm{X} \leq 1.5)$ is
  • A
    $\frac{1}{4}$
  • ✓
    $\frac{1}{2}$
  • C
    $\frac{3}{4}$
  • D
    1
Answer
Correct option: B.
$\frac{1}{2}$
(B)
$\mathrm{P}(0.5 \leq \mathrm{X} \leq 1.5)=\int_{0.5}^{1.5} \mathrm{f}(x) \mathrm{d} x=\int_{0.5}^{1.5} 0.5 x \mathrm{~d} x$
$=0.5\left[\frac{x^2}{2}\right]_{0.5}^{1.5}=\frac{1}{2}$
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MCQ 852 Marks
The p.d.f. of a random variable $X$ is
$\begin{aligned}\mathrm{f}(x) & =2 x, & & 0 \leq x \leq 1 \\& =0, & & \text { otherwise }\end{aligned}$
Then the value of $\mathrm{P}\left(\frac{1}{3}<\mathrm{X}<\frac{1}{2}\right)$ is
  • A
    $\frac{1}{36}$
  • ✓
    $\frac{5}{36}$
  • C
    $\frac{7}{36}$
  • D
    $\frac{11}{36}$
Answer
Correct option: B.
$\frac{5}{36}$
(B)
$\begin{aligned} \mathrm{P} & \left(\frac{1}{3}<\mathrm{X}<\frac{1}{2}\right)=\int_{1 / 3}^{1 / 2} \mathrm{f}(x) \mathrm{d} x \\ & =\int_{1 / 3}^{1 / 2} 2 x \mathrm{~d} x=\left[x^2\right]_{1 / 3}^{1 / 2}=\frac{5}{36}\end{aligned}$
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MCQ 862 Marks
The p.d.f. of a random variable X is
$\begin{aligned}\mathrm{f}(x) & =\frac{1}{5}, & & 0 \leq x \leq 5 \\& =0, & & \text { otherwise }\end{aligned}$
then the value of $\mathrm{P}(1<\mathrm{X}<3)$ is
  • A
    $\frac{1}{5}$
  • ✓
    $\frac{2}{5}$
  • C
    $\frac{3}{5}$
  • D
    $\frac{4}{5}$
Answer
Correct option: B.
$\frac{2}{5}$
(B)
$\mathrm{P}(1<\mathrm{X}<3)=\int_1^3 \mathrm{f}(x) \mathrm{d} x$
$=\int_1^3 \frac{1}{5} \mathrm{~d} x=\frac{1}{5}[x]_1^3=\frac{2}{5}$
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MCQ 872 Marks
$\mathrm{f}(x)=\left\{\begin{array}{cc}\mathrm{k} x^2 ; & 0 \leq x \leq 2 \\0 ; & \text { otherwise }\end{array}\right.$ is a p.d.f. of $X$.
The value of $k$ is
  • A
    $\frac{5}{8}$
  • ✓
    $\frac{3}{8}$
  • C
    $\frac{1}{8}$
  • D
    $\frac{2}{8}$
Answer
Correct option: B.
$\frac{3}{8}$
(B)
Since $\mathrm{f}(x)$ is the p.d.f. of X ,
$\begin{aligned}& \int_{-\infty}^{\infty} \mathrm{f}(x) \mathrm{d} x=1 \\& \Rightarrow \int_0^2 \mathrm{k} x^2 \mathrm{~d} x=1 \Rightarrow \mathrm{k}\left[\frac{x^3}{3}\right]_0^2=1 \\& \Rightarrow \mathrm{k}\left[\frac{8}{3}\right]=1 \Rightarrow \mathrm{k}=\frac{3}{8}\end{aligned}$
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MCQ 882 Marks
A random variable X has the following probability distribution:
X 1 2 3 4
P(X) $\frac{1}{7}$ $\frac{2}{7}$ $\frac{3}{7}$ $\frac{1}{7}$
Then, the variance of this distribution is
  • A
    $\frac{49}{40}$
  • ✓
    $\frac{40}{49}$
  • C
    $\frac{20}{29}$
  • D
    $\frac{29}{20}$
Answer
Correct option: B.
$\frac{40}{49}$
(B)
$\mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}}, \mathrm{P}\left(x_{\mathrm{i}}\right)$
$=1\left(\frac{1}{7}\right)+2\left(\frac{2}{7}\right)+3\left(\frac{3}{7}\right)+4\left(\frac{1}{7}\right)=\frac{18}{7}$
$\begin{aligned} E\left(X^2\right)=\left(1^2\right)\left(\frac{1}{7}\right)+\left(2^2\right)\left(\frac{2}{7}\right)+\left(3^2\right)\left(\frac{3}{7}\right) +\left(4^2\right)\left(\frac{1}{7}\right)\end{aligned}$
$=\frac{1}{7}+\frac{8}{7}+\frac{27}{7}+\frac{16}{7}=\frac{52}{7}$
$\therefore \quad \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2$
$=\frac{52}{7}-\left(\frac{18}{7}\right)^2=\frac{40}{49}$
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MCQ 892 Marks
A random variable X takes values 1,2,3,4 with probabilities $\frac{1}{6}, \frac{1}{3}, \frac{1}{3}, \frac{1}{6}$ respectively, then its mean and variance are respectively
  • ✓
    $\frac{5}{2}, \frac{11}{12}$
  • B
    $\frac{5}{2}, \frac{11}{16}$
  • C
    $\frac{5}{3}, \frac{11}{16}$
  • D
    $\frac{5}{3}, \frac{11}{12}$
Answer
Correct option: A.
$\frac{5}{2}, \frac{11}{12}$
(A)
$\text {Mean}=\mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right)$
$ =\frac{1}{6}(1)+\frac{1}{3}(2)+\frac{1}{3}(3)+\frac{1}{6}(4) $
$ =\frac{1+4+6+4}{6} $
$ =\frac{15}{6}=\frac{5}{2} $
$ \text { Variance }=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 $
$ =\frac{1}{6}(1)^2+\frac{1}{3}(2)^2+\frac{1}{3}(3)^2+\frac{1}{6}(4)^2-\left(\frac{5}{2}\right)^2 $
$ =\frac{1}{6}+\frac{4}{3}+\frac{9}{3}+\frac{16}{6}-\frac{25}{4} $
$ =\frac{2+16+36+32-75}{12} $
$ =\frac{86-75}{12} $
$ =\frac{11}{12}$
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MCQ 902 Marks
For a random variable $X$, if $E(X)=5$ and $\operatorname{Var}(X)=6$, then $E\left(X^2\right)$ is equal to
  • A
    19
  • ✓
    31
  • C
    61
  • D
    11
Answer
Correct option: B.
31
(B)
$\begin{aligned} & \text {Since, } \operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2 \\ & \therefore \quad 6=E\left(X^2\right)-(5)^2 \\ & \therefore E\left(X^2\right)=25+6=31\end{aligned}$
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MCQ 912 Marks
For a random variable $X$, if $\operatorname{Var}(X)=4$ and $\mathrm{E}\left(\mathrm{X}^2\right)=13$, the value of $\mathrm{E}(\mathrm{X})$ is
  • ✓
    3
  • B
    4
  • C
    5
  • D
    6
Answer
Correct option: A.
3
(A)
$\begin{aligned} & \text {Since } \operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2 \\ & \therefore \quad 4=13-[E(X)]^2 \\ & \therefore \quad[E(X)]^2=13-4=9 \\ & \therefore \quad E(X)=3\end{aligned}$
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MCQ 922 Marks
A random variable X has the following probability distribution:
$\mathrm{X}=x_i$0135
P($\mathrm{X}=x$)0.20.50.20.1
Then, the variance of X is
  • A
    2.14
  • B
    2.54
  • C
    2.34
  • ✓
    2.24
Answer
Correct option: D.
2.24
(D)
$\mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right)$
$\begin{aligned} & \quad=0(0.2)+1(0.5)+3(0.2)+5(0.1) -0+0.5+0.6+0.5 \\ & \quad=1.6 \\ & \begin{aligned} \text { Variance } & =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ & =(0)^2(0.2)+(1)^2(0.5)+(3)^2(0.2)+(5)^2(0.1)-(1.6)^2 \\ & =4.8-2.56=2.24\end{aligned}\end{aligned}$
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MCQ 932 Marks
The probability distribution of a random variable X is given below:
$\mathrm{X}=x_i$123
$\mathrm{P}\left(\mathrm{X}=x_i\right)$$\frac{1}{4}$$\frac{1}{8}$$\frac{5}{8}$

Then its mean is
  • ✓
    $\frac{19}{8}$
  • B
    $\frac{5}{4}$
  • C
    1
  • D
    $\frac{4}{5}$
Answer
Correct option: A.
$\frac{19}{8}$
(A)
Mean = (1) $\left(\frac{1}{4}\right)+(2)\left(\frac{1}{8}\right)+(3)\left(\frac{5}{8}\right)-\frac{19}{8}$
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MCQ 942 Marks
The probability distribution of a r.v. $X$ is
X-1.5-0.50.51.52.5
P($\mathrm{X}=\boldsymbol{x}$)0.050.20.150.250.35

Then the c.d.f. of X is
  • A
    X-1.5-0.50.51.52.5
    F(X)0.050.20.30.40.6
  • B
    X-1.5-0.50.51.52.5
    F(X)0.050.20.250.350.95
  • C
    X-1.5-0.50.51.52.5
    F(X)0.050.40.450.550.75
  • ✓
    X-1.5-0.50.51.52.5
    F(X)0.050.250.40.650.1
Answer
Correct option: D.
X-1.5-0.50.51.52.5
F(X)0.050.250.40.650.1
(D)
$\begin{aligned} & \mathrm{F}\left(x_1\right)=\mathrm{p}_1=0.05 \\ & \mathrm{~F}\left(x_2\right)=\mathrm{p}_1+\mathrm{p}_2=0.05+0.2=0.25 \\ & \mathrm{~F}\left(x_3\right)=\mathrm{p}_1+\mathrm{p}_2+\mathrm{p}_3=0.25+0.15=0.4 \\ & \mathrm{~F}\left(x_4\right)=\mathrm{p}_1+\mathrm{p}_2+\mathrm{p}_3+\mathrm{p}_4=0.4+0.25=0.65 \\ & \mathrm{~F}\left(x_5\right)=\mathrm{p}_1+\mathrm{p}_2+\mathrm{p}_3+\mathrm{p}_4+\mathrm{p}_5=0.65+0.35=1\end{aligned}$
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MCQ 952 Marks
The probability distribution of a r.v. $X$ is
$\mathrm{X}=\boldsymbol{x}$-2-1012
P($\mathrm{X}=\boldsymbol{x}$)0.20.30.150.250.1
Then F(-1) =
  • ✓
    0.5
  • B
    0.7
  • C
    0.9
  • D
    1
Answer
Correct option: A.
0.5
(A)
F (-1) = 0.2 +0.3 = 0.5
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MCQ 962 Marks
The p.m.f. of a r.v. X is given by
$\mathrm{P}(\mathrm{X}=x)=\frac{{ }^5 \mathrm{C}_x}{2^5}, \quad x=0,1,2, \ldots, 5$
$=0, \quad$ otherwise
Then, $\mathrm{P}(\mathrm{X} \leq 2)=$
  • A
    $\frac{3}{32}$
  • B
    $\frac{7}{32}$
  • C
    $\frac{11}{32}$
  • ✓
    $\frac{16}{32}$
Answer
Correct option: D.
$\frac{16}{32}$
(D)
$ \mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)$
$\begin{aligned} & =\frac{{ }^5 \mathrm{C}_0}{2^5}+\frac{{ }^5 \mathrm{C}_1}{2^5}+\frac{{ }^5 \mathrm{C}_2}{2^5} \\ & =\frac{1+5+10}{2^5}=\frac{16}{32}\end{aligned}$
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MCQ 972 Marks
Let X denote the number of fruits that are grown in a farm house in a particular day. The probability that X can take the value of $x$ has the following form, where k is some unknown constant.
$\mathrm{P}(\mathrm{X}=x)= \begin{cases}\mathrm{k}, & \text { if } x=0 \\ 2 \mathrm{k}, & \text { if } x=1 \\ 3 \mathrm{k}, & \text { if } x=2 \\ 0, & \text { otherwise }\end{cases}$
Then the value of $k$ is
  • A
    $\frac{1}{8}$
  • B
    $\frac{1}{2}$
  • C
    $\frac{1}{3}$
  • ✓
    $\frac{1}{6}$
Answer
Correct option: D.
$\frac{1}{6}$
(D)
The probability distribution of X is
X012
P (X)k2k3k

$ Since \sum_{x=0}^2 \mathrm{P}(\mathrm{X}=x)-1,$
$\begin{aligned} & \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}=1 \\ & \Rightarrow 6 \mathrm{k}=1 \Rightarrow \mathrm{k}=\frac{1}{6}\end{aligned}$
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MCQ 982 Marks
The following table represents a probability distribution for a random variable X :
$\mathrm{X}=\boldsymbol{x}$01234
P($\mathrm{X}=\boldsymbol{x}$)k2k3k4k5k
Then the value of k is
  • A
    $\frac{1}{2}$
  • B
    $\frac{1}{6}$
  • C
    $\frac{1}{3}$
  • ✓
    $\frac{1}{9}$
Answer
Correct option: D.
$\frac{1}{9}$
(D)
$Since \sum_{x=0}^4 \mathrm{P}(\mathrm{X}=x)=1$
$\begin{aligned} & \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}+2 \mathrm{k}+\mathrm{k}=1 \\ & \Rightarrow 9 \mathrm{k}=1 \Rightarrow \mathrm{k}=\frac{1}{9}\end{aligned}$
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MCQ 992 Marks
A random variable $X$ has the following probability distribution:
$\mathrm{X}=\boldsymbol{x}$1234
P($\mathrm{X}=\boldsymbol{x}$)$\frac{1}{8}$$\frac{1}{2}$$\frac{1}{4}$k
Then the value of k is
  • A
    $\frac{1}{4}$
  • ✓
    $\frac{1}{8}$
  • C
    $\frac{3}{8}$
  • D
    $\frac{1}{2}$
Answer
Correct option: B.
$\frac{1}{8}$
(B)
$Since \sum_{x=1}^4 \mathrm{P}(\mathrm{X}=x)=1,$
$\begin{aligned} & \frac{1}{8}+\frac{1}{2}+\frac{1}{4}+\mathrm{k}=1 \\ & \Rightarrow \mathrm{k}+\frac{1+4+2}{8}=1 \\ & \Rightarrow \mathrm{k}=1-\frac{7}{8}=\frac{1}{8}\end{aligned}$
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MCQ 1002 Marks
The probability distribution of X is
X0123
P(X)0.3k2k2k
The value of k is
  • ✓
    0.14
  • B
    0.3
  • C
    0.7
  • D
    1
Answer
Correct option: A.
0.14
(A)
$Since \sum_{x=1}^3 \mathrm{P}(\mathrm{X}=x)=1,$
$\begin{aligned} & 0.3+\mathrm{k}+2 \mathrm{k}+2 \mathrm{k}=1 \\ & \Rightarrow 5 \mathrm{k}=0.7 \\ & \Rightarrow \mathrm{k}=0.14\end{aligned}$
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