M.C.Q (1 Marks)

MCQ 11 Mark

The probability distribution of a discrete random variable $X$ is given below:

$\text{X}:$	$1$	$2$	$3$	$4$
$\text{P}(\text{X}):$	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$

The value of $E(X^2)$ is:

A
$3$
B
$5$
C
$7$
✓
$10$

Answer

Correct option: D.

$10$

$\text{X}$	$1$	$2$	$3$	$4$
$\text{P}(\text{X})$	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$
$\text{X}^2\text{P(X)}$	$\frac{1}{10}$	$\frac{4}{5}$	$\frac{27}{10}$	$\frac{32}{5}$	$\text{E}(\text{X}^2)=10$

View full question & answer→

MCQ 21 Mark

The probability distribution of a discrete random variable X is given below:

$\text{X}:$	$2$	$3$	$4$	$5$
$\text{P}(\text{X}):$	$\frac{5}{\text{k}}$	$\frac{7}{\text{k}}$	$\frac{9}{\text{k}}$	$\frac{11}{\text{k}}$

The value of k is:

A
8
B
16
✓
32
D
48

Answer

Correct option: C.

32

$\sum\limits_2^5\text{P}(\text{x})=1$
$\frac{5}{\text{k}}+\frac{7}{\text{k}}+\frac{9}{\text{k}}+\frac{11}{\text{k}}=1$

$\text{k}=32$

NOTE: Question is modified.

View full question & answer→

MCQ 31 Mark

Let $X$ be a discrete random variable. Then the variance of $X$ is:

A
$E(X^2)$
B
$E(X^2) + (E(X))^2$
✓
$E(X^2) - (E(X))^2$
D
$\sqrt{\text{E}(\text{X}^2)-(\text{E}(\text{X}))^2}$

Answer

Correct option: C.

$E(X^2) - (E(X))^2$

Since, the variance of a discrete random variable $X$ is given by :
Var $(X) = E(X^2) - (E(X))^2$
Hence, the correct alternative is option $(c).$

View full question & answer→

MCQ 41 Mark

For the following probability distribution:

X:	-4	-3	-2	-1	0
P(X):	0.1	0.2	0.3	0.2	0.2

The value of E(X) is:

A
0
B
-1
C
-2
✓
-1.8

Answer

Correct option: D.

-1.8

The probability distribution of X is given below:

X:	-4	-3	-2	-1	0
P(X):	0.1	0.2	0.3	0.2	0.2

E(X) = (-4) × 0.1 + (-3) × 0.2 + (-2) × 0.3 + (-1) × 0.2 + 0 × 0.2

= -0.4 - 0.6 - 0.6 - 0.2

= -1.8

Hence, the correct alternative is option (d).

View full question & answer→

MCQ 51 Mark

If $X$ is a random variable with probability distribution as given below :

$X = x_i$	$0$	$1$	$2$	$3$
$P(X = X_i)$	$k$	$3k$	$3k$	$k$

The value of $k$ and its variance are :

A
$\frac{1}{8},\frac{22}{27}$
B
$\frac{1}{8},\frac{23}{27}$
C
$\frac{1}{8},\frac{24}{27}$
✓
$\frac{1}{8},\frac{3}{4}$

Answer

Correct option: D.

$\frac{1}{8},\frac{3}{4}$

$\sum\limits_0^3\text{P}(\text{x})=1$
$\text{k}+3\text{k}+3\text{k}+\text{k}=1$
$\text{k}=\frac{1}{8}$

$\text{x}$	$\text{P}(\text{x})$	$\text{x}\text{P}(\text{x})$	$\text{x}^2\text{P}(\text{x})$
$0$	$\frac{1}{8}$	$0$	$0$
$1$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{3}{8}$
$2$	$\frac{3}{8}$	$\frac{6}{8}$	$\frac{12}{8}$
$3$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{9}{8}$
$\text{Total}$		$\text{E(x)}=\frac{12}{8}=1.5$	$\text{E}(\text{x}^2)=3$

$\text{V(x)}=\text{E}(\text{x}^2)-[\text{E}(\text{x})^2]$
$\text{V(x)}=3-(1.5)^2$
$\text{V(x)}=0.75=\frac{3}{4}$

View full question & answer→

MCQ 61 Mark

If the random variable X has the following distribution:

X:	0	1	2	3	4	5	6	7	8
P(X):	a	3a	5a	7a	9a	11a	13a	15a	17a

then the value of a is:

A
$\frac{7}{81}$
B
$\frac{5}{81}$
C
$\frac{2}{81}$
✓
$\frac{1}{81}$

Answer

Correct option: D.

$\frac{1}{81}$

We know that the sum of probsabilities in a probability distribution is always 1.
$\therefore$ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 1

⇒ a + 3a+ 5a+ 7a+ 9a + 11a + 13a + 15a + 17a = 1

⇒ 81a = 1

$\Rightarrow\text{a}=\frac{1}{81}$

View full question & answer→

MCQ 71 Mark

A random variable $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3$. If $P(X = 3) = 2P(X = 1)$ and $P(X = 2) = 0.3,$ then $P(X = 0)$ is :

A
$0.1$
B
$0.2$
C
$0.3$
✓
$0.4$

Answer

Correct option: D.

$0.4$

Let : $P(X = 0) = m$
$P(X = 1) = k$
Now, $P(X = 3) = 2k$

$x_i$	$p_i$	$p_ix_i$
$0$	$m$	$0$
$1$	$k$	$k$
$2$	$0.3$	$0.6$
$3$	$2k$	$6k$

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}$
$\Rightarrow 0 + k + 0.6 + 6k = 1.3$
$\Rightarrow 7k = 1.6 - 0.6$
$\Rightarrow\text{k}=\frac{0.7}{7}$
$\Rightarrow 0.1$
We know that the sum of probabilities in a probability distribution is always $1$.
$\therefore P(X = 0) + P(X = 1) + P(X = 3) = 1$
$\Rightarrow m + 0.1 + 0.3 + 0.2 = 1$
$\Rightarrow m + 0.6 = 1$
$\Rightarrow m = 0.4$

View full question & answer→

MCQ 81 Mark

A random variable has the following probability distribution:

$X = x_i$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X = X_i)$	$0$	$2p$	$2p$	$3p$	$p^2$	$2p^2$	$7p^2$	$7p^2$	$2p$

✓
$\frac{1}{10}$
B
$-1$
C
$-\frac{1}{10}$
D
$\frac{1}{5}$

Answer

Correct option: A.

$\frac{1}{10}$

We know that the sum of probabilities in a probability distribution is always $1$.
$\therefore P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) +P(X = 6) + P(X = 7) +P(X = 8) = 1$
$\Rightarrow 0 + 2p + 2p + 3p + p^2 + 2p^2 + 7p^2 + 2p = 1$
$\Rightarrow 10p^2+ 9p - 1 = 0$
$\Rightarrow (10p - 1)(p + 1) = 0$
$\Rightarrow\text{p}=\frac{1}{10}\text{ or }-1 \ ($Negleting $-1$ as the value of the probability cannot be negative$)$

View full question & answer→

MCQ 91 Mark

A random variable X has the following probability distribution:

X:	1	2	3	4	5	6	7	8
P(X):	0.15	0.23	0.12	0.10	0.20	0.08	0.07	0.05

Find the events E = {X : X is a prime number}, F{X : X < 4}, the probability $\text{P}(\text{E}\cup\text{F})$ is:

✓
0.50
B
0.77
C
0.35
D
0.87

Answer

Correct option: A.

0.50

P(E) = P(2) + P(3) + P(5) + P(7)
P(E) = 0.23 + 0.12 + 0.20 + 0.07

P(E) = 0.62

And

P(F) = P(1) + P(2) + P(3)

P(F) = 0.15 + 0.23 + 0.12

P(F) = 0.5

Also,

$\text{P}(\text{E}\cap\text{F})=\text{P}(2)+\text{P}(3)$

$\text{P}(\text{E}\cap\text{F})=0.23+0.12$

$\text{P}(\text{E}\cap\text{F})=0.35$

$\text{P}(\text{E}\cup\text{F})=\text{P}(\text{E})+\text{P(F)}-\text{P}(\text{E}\cap\text{F})$

$\text{P}(\text{E}\cup\text{F})=0.62+0.5-0.35$

$\text{P}(\text{E}\cup\text{F})=0.77$

View full question & answer→

MATHS M.C.Q (1 Marks)

Test yourself on this topic