Step-by-Step Solution

Topic: Probability Distributions & Expectation

(I) DETERMINE THE VALUE OF K

The sum of probabilities in a probability distribution must equal 1.

\( \sum P(X=x) = 1 \)

Substituting the values for \( x = 0, 1, 2, 3, 4 \):

\( 0 + \frac{k(1)}{6} + \frac{k(2)}{6} + \frac{(1-k)3}{6} + \frac{k(4)}{2} = 1 \)

Simplify the terms:

\( \frac{3k}{6} + \frac{3(1-k)}{6} + 2k = 1 \)   ⇒   \( \frac{3k + 3 – 3k}{6} + 2k = 1 \)

\( \frac{3}{6} + 2k = 1 \)   ⇒   \( 0.5 + 2k = 1 \)

2k = 0.5   ⇒   k = 0.25

(II) PROBABILITY GIVEN X ≥ 3

We need to find the probability that a student applies at least 3 weeks ahead. Since the distribution describes admitted students, this is \( P(X \ge 3) \).

First, calculate individual probabilities using \( k = 0.25 \):

  • \( P(3) = \frac{(1 – 0.25) \times 3}{6} = \frac{0.75 \times 3}{6} = \frac{2.25}{6} = 0.375 \)
  • \( P(4) = \frac{0.25 \times 4}{2} = 0.5 \)

\( P(X \ge 3) = P(3) + P(4) = 0.375 + 0.5 \)

Probability = 0.875 (or 7/8)

(III)(A) MATHEMATICAL EXPECTATION E(X)

Expectation \( E(X) = \sum x \cdot P(x) \):

  • \( x=0 \): \( 0 \)
  • \( x=1 \): \( 1 \times \frac{0.25}{6} = \frac{1}{24} \)
  • \( x=2 \): \( 2 \times \frac{0.5}{6} = \frac{2}{12} = \frac{4}{24} \)
  • \( x=3 \): \( 3 \times 0.375 = 1.125 = \frac{27}{24} \)
  • \( x=4 \): \( 4 \times 0.5 = 2.0 = \frac{48}{24} \)

Sum = \( \frac{1 + 4 + 27 + 48}{24} = \frac{80}{24} \)

E(X) = 10/3 ≈ 3.33 weeks

(III)(B) EXPECTED SCHOLARSHIP

Multiply the scholarship amount by the probability of applying in that week:

  • 4 weeks: \( 50,000 \times 0.5 = 25,000 \)
  • 3 weeks: \( 20,000 \times 0.375 = 7,500 \)
  • 2 weeks: \( 12,000 \times (1/12) = 1,000 \)
  • 1 week: \( 9,600 \times (1/24) = 400 \)
Expected Scholarship = ₹ 33,900
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