Search Previous Year Questions

Calculation:

Given,

The probability that A , B , and C  become managers are:

$P(A) = \frac{3}{10}, , P(B) = \frac{1}{2}, , P(C) = \frac{4}{5}$

The conditional probabilities that the bonus scheme is introduced are:

$P(D|A) = \frac{4}{9}, , P(D|B) = \frac{2}{9}, , P(D|C) = \frac{1}{3}$

The total probability that the bonus scheme will be introduced is given by:

$P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)$

Substituting the values:

$P(D) = \left(\frac{4}{9} \times \frac{3}{10}\right) + \left(\frac{2}{9} \times \frac{1}{2}\right) + \left(\frac{1}{3} \times \frac{4}{5}\right)$

Now, simplifying the terms:

$P(D) = \frac{12}{90} + \frac{2}{18} + \frac{4}{15}$

The least common denominator (LCD) is 90. So:

$P(D) = \frac{12}{90} + \frac{10}{90} + \frac{24}{90}$

Adding the fractions gives:

$P(D) = \frac{46}{90} = \frac{23}{45}$

Hence, the correct answer is Option 3.

Calculation:

Given,

The probability that A , B , and C  become managers are:

$P(A) = \frac{3}{10}, , P(B) = \frac{1}{2}, , P(C) = \frac{4}{5}$

The conditional probabilities that the bonus scheme is introduced are:

$P(D|A) = \frac{4}{9}, , P(D|B) = \frac{2}{9}, , P(D|C) = \frac{1}{3}$

We need to find the probability that B  is the manager given that the bonus scheme has been introduced using Bayes' Theorem:

$P(B|D) = \frac{P(D|B)P(B)}{P(D)}$

First, calculate the total probability P(D)  that the bonus scheme is introduced:

$P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)$

Substitute the values:

$P(D) = \left( \frac{4}{9} \times \frac{3}{10} \right) + \left( \frac{2}{9} \times \frac{1}{2} \right) + \left( \frac{1}{3} \times \frac{4}{5} \right)$

Simplifying:

$P(D) = \frac{12}{90} + \frac{2}{18} + \frac{4}{15}$

Find a common denominator (LCD = 90):

$P(D) = \frac{12}{90} + \frac{10}{90} + \frac{24}{90} = \frac{46}{90} = \frac{23}{45}$

Now, use Bayes' Theorem to find P(B|D) :

$P(B|D) = \frac{P(D|B)P(B)}{P(D)} = \frac{\left( \frac{2}{9} \times \frac{1}{2} \right)}{\frac{23}{45}}$

Simplifying:

$P(B|D) = \frac{\frac{2}{18}}{\frac{23}{45}} = \frac{\frac{1}{9}}{\frac{23}{45}} = \frac{5}{23}$

The probability that the manager appointed was B , given that the bonus scheme has been introduced, is 5/23.

Hence, the correct answer is Option 1.

Calculation:

Given,

Number of observations, n = 100

Original arithmetic mean, $\bar{x} = 50$

Transformation applied to each observation:

New value, $x' = \frac{x - 5}{20}$

The new arithmetic mean is,

$\bar{x}' = \frac{\bar{x} - 5}{20}$

$\bar{x}' = \frac{50 - 5}{20} = \frac{45}{20} = 2.25$

∴ The new arithmetic mean is 2.25.

Hence, the correct answer is Option 1.

Calculation:

Given,

Number of observations, n = 100

Original standard deviation, $\sigma = 10$

Transformation applied to each observation:

New value, $x' = \frac{x + 5}{20}$

The additive constant (5) does not affect the standard deviation, and scaling by 20 divides the standard deviation by 20, so the new standard deviation is,

$\sigma' = \frac{\sigma}{20}$

$\sigma' = \frac{10}{20} = 0.5$

∴ The new standard deviation is 0.5.

Hence, the correct answer is Option 2.

Calculation:

Given,

P(A) = 1/3

P(B) = 1/2

P(A ∩ B) = 1/4

We need to calculate the probability P(B | A^C), the probability that B occurs given that A does not occur.

Using the formula for conditional probability:

P(B | A^C) = P(B ∩ A^C) / P(A^C)

Now,

P(A^C) = 1 - P(A) = 1 - 1/3 = 2/3

P(B ∩ A^C) = P(B) - P(A ∩ B) = 1/2 - 1/4 = 1/4

Substitute the values:

P(B | A^C) = (1/4) / (2/3) = (1/4) × (3/2) = 3/8

Hence, the correct answer is Option 2.

Calculation:

Given,

$P(A) = \frac{1}{3}$

$P(B) = \frac{1}{2}$

$P(A \cap B) = \frac{1}{4}$

We need to calculate$P(A^C \cap B^C)$, the probability that neither A  nor B  occurs.

Using the complement rule:

$P(A^C \cap B^C) = 1 - P(A \cup B)$

Now, apply the inclusion-exclusion formula to find $( P(A \cup B)$:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute the given values:

$P(A \cup B) = \frac{1}{3} + \frac{1}{2} - \frac{1}{4}$

To simplify:

$P(A \cup B) = \frac{4}{12} + \frac{6}{12} - \frac{3}{12} = \frac{7}{12}$

Now, calculate $P(A^C \cap B^C)$:

$P(A^C \cap B^C) = 1 - \frac{7}{12} = \frac{5}{12}$

Hence, the correct answer is Option 2.

Calculation:

Given,

Two fair six-sided dice are rolled.

Each die has faces numbered 1 through 6.

Total number of outcomes:

Since each die has 6 faces, the sample space size is

$N = 6 \times 6 = 36$.

Favourable outcomes (sum > 7):

We list all ordered pairs ((i, j)) with (i + j > 7):

Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes

Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes

Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes

Sum = 11: (5,6), (6,5) → 2 outcomes

Sum = 12: (6,6) → 1 outcome

Total favourable outcomes = (5 + 4 + 3 + 2 + 1 = 15).

Probability calculation:

The probability that the sum is strictly greater than 7 is

$P(\text{sum} > 7) = \frac{\text{favourable outcomes}}{\text{total outcomes}} = \frac{15}{36} = \frac{5}{12}$.

Hence, the correct answer is Option 2.

Calculation:

Given,

Probability of hitting the target per shot, $p = \frac{1}{5}$

Number of shots fired, $n = 7$

The number of hits follows a binomial distribution:

$X \sim \mathrm{Binomial}(n=7,;p=\tfrac15)$

Probability of zero hits:

$P(X=0)=\left(\frac45\right)^{7}$

Probability of exactly one hit:

$P(X=1)=\binom71!\left(\frac15\right)!\left(\frac45\right)^{6} =\frac{7}{5}\left(\frac45\right)^{6}$

Probability of at least two hits:

$P(X\ge2)=1-\bigl[P(X=0)+P(X=1)\bigr] =1-\frac{11}{5}\left(\frac45\right)^{6}$

∴ The probability of hitting the target at least twice is  $1-\dfrac{11}{5}\left(\dfrac45\right)^{6}$.

Hence, the correct answer is Option 3.

Calculation:

Given,

Mean of the binomial distribution, $\mu = 200$

Variance of the binomial distribution, $\sigma^{2} = 160$

For a binomial random variable,

$\mu = np \quad\text{and}\quad \sigma^{2} = np(1-p).$

From the mean,

$np = 200 ;\Longrightarrow; p = \dfrac{200}{n}.$

Using the variance,

$np(1-p) = 160 ;\Longrightarrow; n\bigl(\tfrac{200}{n}\bigr)!\Bigl(1-\tfrac{200}{n}\Bigr)=160 ;\Longrightarrow; 200\Bigl(1-\tfrac{200}{n}\Bigr)=160.$

Simplifying,

$1-\tfrac{200}{n} = 0.8 ;\Longrightarrow; \tfrac{200}{n} = 0.2 ;\Longrightarrow; n = \dfrac{200}{0.2} = 1000.$

∴ The number of trials is $n = 1000$.

Calculation:

Given,

The sequence is $8^{2},,9^{2},,10^{2},,\dots,,15^{2}$

Number of terms, $n = 15 - 8 + 1 = 8$

The sum of the first $m$ squares is  $\displaystyle \sum_{k=1}^{m} k^{2} = \frac{m(m+1)(2m+1)}{6}$

Sum up to $15^{2}$:

$S_{15} = \frac{15 \times 16 \times 31}{6} = 1240$

Sum up to $7^{2}$:

$S_{7} = \frac{7 \times 8 \times 15}{6} = 140$

Sum of the required terms:

$S = S_{15} - S_{7} = 1240 - 140 = 1100$

The arithmetic mean is,

$\displaystyle \text{Mean} = \frac{S}{n} = \frac{1100}{8} = 137.5$

∴ The arithmetic mean is $137.5$.

Hence, the correct answer is option 3.