Direction: Given below are two quantities named I and II. Based on the

1.	Direction: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: A bag contains 5 red, 3 black, 4 yellow, and 4 pink pens. If 4 pens are drawn at random, find the probability that 2 pens are red and 2 pens are black.Quantity II: Find the probability such that all 4 jacks are drawn when 5 cards are drawn at random from a pack of 52 cards.1. Quantity I ≤ Quantity II2. Quantity I ≥ Quantity II3. Quantity I ˃ Quantity II4. Quantity I ˂ Quantity II5. Quantity I = Quantity II
Answer» Correct Answer - Option 3 : Quantity I ˃ Quantity II Quantity I: Total number of outcomes = ¹⁶C₄ ⇒ 16!/(4! × 12!) ⇒ (16 × 15 × 14 × 13 × 12!) / (4 × 3 × 2 × 1 × 12!) ⇒ 1820 Number of favourable outcomes = ⁵C₂ × ³C₂ ⇒ 30 ⇒ Required probability = 30/1820 ⇒ 0.0165 (approx) Quantity II: Total number of possible ways to drawn a card = ⁵²C₅ ⇒ 52!/(5! × 47!) ⇒ (52 × 51 × 50 × 49 × 48 × 47!)/ (5 × 4 × 3 × 2 × 1 × 47!) ⇒ 2598960 Number of sets of 5 with all are jack = ⁴C₄ × ⁴⁸C₁ ⇒ 1 × (48 × 47!) / (1! × 47!) = 48 ⇒ P(E ) = 48/2598960 ⇒ 1/54145 ⇒ 0.000018(approx) ∴ Quantity I > Quantity II

Direction: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: A bag contains 5 red, 3 black, 4 yellow, and 4 pink pens. If 4 pens are drawn at random, find the probability that 2 pens are red and 2 pens are black.Quantity II: Find the probability such that all 4 jacks are drawn when 5 cards are drawn at random from a pack of 52 cards.1. Quantity I ≤ Quantity II2. Quantity I ≥ Quantity II3. Quantity I ˃ Quantity II4. Quantity I ˂ Quantity II5. Quantity I = Quantity II

Answer» Correct Answer - Option 3 : Quantity I ˃ Quantity II

Quantity I:

Total number of outcomes = ¹⁶C₄

⇒ 16!/(4! × 12!)

⇒ (16 × 15 × 14 × 13 × 12!) / (4 × 3 × 2 × 1 × 12!)

⇒ 1820

Number of favourable outcomes = ⁵C₂ × ³C₂

⇒ 30

⇒ Required probability = 30/1820

⇒ 0.0165 (approx)

Quantity II:

Total number of possible ways to drawn a card = ⁵²C₅

⇒ 52!/(5! × 47!)

⇒ (52 × 51 × 50 × 49 × 48 × 47!)/ (5 × 4 × 3 × 2 × 1 × 47!)

⇒ 2598960

Number of sets of 5 with all are jack = ⁴C₄ × ⁴⁸C₁

⇒ 1 × (48 × 47!) / (1! × 47!) = 48

⇒ P(E ) = 48/2598960

⇒ 1/54145

⇒ 0.000018(approx)

∴ Quantity I > Quantity II