NIMCET 2008 — Mathematics PYQ

Q: How do I solve NIMCET 2008 Mathematics questions?

Practice NIMCET 2008 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of NIMCET Mathematics questions?

NIMCET Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

NIMCET 2008 — Mathematics PYQ

NIMCET | Mathematics | 2008

A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters, TA, are visible. The probability that the letter has come from CALCUTTA is:

Choose the correct answer:

Explanation

Step 1: Define the Events

Let $E_1$ : The letter came from TATANAGAR.
Let $E_2$ : The letter came from CALCUTTA.
Let $A$ : The visible consecutive letters are TA.

Since it is equally likely for the letter to come from either city:

P(E_1) = P(E_2) = \frac{1}{2}

Step 2: Find the number of consecutive pairs in each word

To find the probability of seeing "TA", we count all possible pairs of consecutive letters.

For TATANAGAR:

The pairs are: TA, AT, TA, AN, NA, AG, GA, AR.

Total pairs = $8$

Number of "TA" pairs = $2$ (at positions 1-2 and 3-4)

$P(A|E_1) = \frac{2}{8} = \frac{1}{4}$
For CALCUTTA:

The pairs are: CA, AL, LC, CU, UT, TT, TA.

Total pairs = $7$

Number of "TA" pairs = $1$ (at positions 7-8)

$P(A|E_2) = \frac{1}{7}$

Step 3: Apply Bayes' Theorem

We need to find the probability that the letter came from CALCUTTA given that "TA" is visible ( $P(E_2|A)$ ):

P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}

Substitute the values:

P(E_2|A) = \frac{\frac{1}{2} \cdot \frac{1}{7}}{\left(\frac{1}{2} \cdot \frac{1}{4}\right) + \left(\frac{1}{2} \cdot \frac{1}{7}\right)}

Cancel the common factor of $\frac{1}{2}$ :

P(E_2|A) = \frac{\frac{1}{7}}{\frac{1}{4} + \frac{1}{7}}

Step 4: Simplify the fraction

Calculate the denominator: $\frac{1}{4} + \frac{1}{7} = \frac{7 + 4}{28} = \frac{11}{28}$

P(E_2|A) = \frac{1/7}{11/28}

P(E_2|A) = \frac{1}{7} \times \frac{28}{11} = \frac{4}{11}

Conclusion:

The probability that the letter has come from CALCUTTA is 4/11.

Correct Option: (a)

Choose the correct answer:

Explore More