Search Previous Year Questions

Calculation:

Total number of outcomes for 4 coin tosses:

$n(S) = 2^4 = 16$

Favorable outcomes with two consecutive heads:

⇒ $E = { \text{HHTT, THHT, TTHH, HHHT, HTHH, HHTH, THHH, HHHH} }$

⇒ $n(E) = 8 \Rightarrow n(E') = 16 - 8 = 8$

Probability of not getting two consecutive heads:

⇒ $P(E') = \frac{8}{16} = \frac{1}{2}$

Hence, the correct answer is Option 4.

Calculation:

Given:

Prime numbers = {2, 3, 5} ⇒ 3 choices

Composite numbers = {4, 6} ⇒ 2 choices

Neither prime nor composite = {1} ⇒ 1 choice

Probability of selecting one prime, one composite, and one neither prime nor composite:

$P = \frac{3}{6} \cdot \frac{2}{6} \cdot \frac{1}{6} \cdot 3!$

Calculating:

$= \frac{6}{216} \cdot 6 = \frac{36}{216} = \frac{1}{6}$

Hence, the correct answer is Option 4.

Calculation:

Given:

Numbers divisible by 5 = 10

Numbers divisible by 9 = 5

Numbers divisible by both 5 and 9 = 1

Using the inclusion-exclusion principle:

Divisible by 5 or 9 $= 10 + 5 - 1 = 14$

Total numbers = 50

Not divisible by 5 or 9 $= 50 - 14 = 36$

Required probability:

$P = \frac{36}{50} = \frac{18}{25}$

Hence, the correct answer is Option 2.

Calculation:

Out of 50 consecutive natural numbers:

Even numbers = 25

Odd numbers = 25

Sum of one even and one odd number is always odd.

Number of ways to choose one even and one odd:

$= 25 \cdot 25 = 625$

Total number of ways to choose any two numbers from 50:

$= {}^{50}C_2 = \frac{50 \cdot 49}{2} = 1225$

Required probability:

$P(\text{sum is odd}) = \frac{625}{1225} = \frac{25}{49}$

Hence, the Correct answer is Option 4.

Calculation:

Given:

Standard deviation of 100 observations = 10

A constant value $20$ is added to each observation.

Property of standard deviation:

Adding or subtracting a constant from each observation does not change the standard deviation

Hence,

New standard deviation = Old standard deviation = 10

Hence, the correct answer is Option 1.

Calculation:

Given:

$n = 5$

$p = k$, $q = 1 - k$

Binomial probabilities:

⇒ $P(X = 1) = {}^5C_1 \cdot k \cdot (1 - k)^4 = 0.4096$

⇒ $P(X = 2) = {}^5C_2 \cdot k^2 \cdot (1 - k)^3 = 0.2048$

Taking ratio:

⇒ $\frac{{}^5C_2 \cdot k^2 \cdot (1 - k)^3}{{}^5C_1 \cdot k \cdot (1 - k)^4} = \frac{0.2048}{0.4096}$

Simplifying:

⇒ $\frac{10 \cdot k}{5 \cdot (1 - k)} = \frac{1}{2} \Rightarrow 4k = 1 - k \Rightarrow 5k = 1 \Rightarrow k = \frac{1}{5} = 0.2$

Hence, the correct answer is Option 1.

Calculation:

The given data is a grouped frequency distribution.

Compute the totals:

⇒ $\sum f_i = 20+30+30+20 = 100$

⇒ $\sum x_i f_i = 200+600+900+800 = 2500$

Mean of the distribution is given by:

⇒ $\text{Mean} = \frac{\sum x_i f_i}{\sum f_i}$

Substituting the values:

⇒ $\text{Mean} = \frac{2500}{100} = 25$

Hence, the correct answer is Option 2.

Calculation:

Given:

 P(A) = 0.3 , P(B) = 0.4 and P( A |B)=0.5

Using the formula of conditional probability:

$P!\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P(B)}$

Substitute the given values:

$0.5=\frac{P(A\cap B)}{0.4}$

Solving for $P(A\cap B)$:

$P(A\cap B)=0.5\times 0.4=0.2$

Now find $P!\left(\frac{B}{A}\right)$:

$P!\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P(A)}$

Substitute the values:

$P!\left(\frac{B}{A}\right)=\frac{0.2}{0.3}=\frac{2}{3}\approx 0.667$

Hence, the correct answer is Option 4.

Calculation:

Given:

$P(A)=\frac{1}{3},\quad P(B)=\frac{1}{2},\quad P(A\cap B)=\frac{1}{4}$

We know that:

⇒ $P(\bar{A})=1-P(A)$

Using the formula:

⇒ $P(\bar{A}\cup B)=P(\bar{A})+P(B)-P(\bar{A}\cap B)$

Substitute $P(\bar{A})=1-P(A)$:

⇒ $P(\bar{A}\cup B)=1-P(A)+P(B)-[P(B)-P(A\cap B)]$

Substitute the given values:

$=1-\frac{1}{3}+\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{4}\right)$

Simplifying:

$=\frac{2}{3}+\frac{1}{2}-\frac{1}{4}$

$=\frac{8+6-3}{12}=\frac{11}{12}$

Hence, the correct answer is Option 4.

Calculation:

Statement I: Mean and variance have the same unit of measurement.

• Mean has the same unit as the original data.
• Variance is the square of the unit of the data (since deviations are squared).
• Therefore, mean and variance do NOT have the same unit.
• Statement I is incorrect.

Statement II: Mean deviation and standard deviation have the same unit of measurement.

• Mean deviation is based on absolute deviations, so it has the same unit as the data.
• Standard deviation is the square root of variance, hence it also has the same unit as the data.
• Therefore, Statement II is correct.

Hence, the correct answer is Option 2.