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Home/ Questions/Q 343614
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Zeeshan Biren Deep
Zeeshan Biren Deep
Asked: 3 years ago2022-11-08T05:46:13+05:30 2022-11-08T05:46:13+05:30In: Probability

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.

There are 50 cards in a box numbered from 1 to 50. One number is written on each card.A card is drawn from the box.

Quantity I: Find the probability that the drawn card is a multiple of 2, 3 and 4.

Quantity II: Find the probability that the drawn card is an odd prime number.


1. Quantity I ˂ Quantity II
2. Quantity I ≥ Quantity II
3. Quantity I ≤ Quantity II
4. Quantity I = Quantity II
5. 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.

There are 50 cards in a box numbered from 1 to 50. One number is written on each card.A card is drawn from the box.

Quantity I: Find the probability that the drawn card is a multiple of 2, 3 and 4.

Quantity II: Find the probability that the drawn card is an odd prime number.


1. Quantity I ˂ Quantity II
2. Quantity I ≥ Quantity II
3. Quantity I ≤ Quantity II
4. Quantity I = Quantity II
5. Quantity I ˃ Quantity II
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  1. df7e4
    2022-11-03T17:59:50+05:30Added an answer about 3 years ago

    Correct Answer – Option 1 : Quantity I ˂ Quantity II

    Quantity I:

    Total number of cards = 50

    Number of favorable cards = 4 (i.e. 12, 24,36 and 48)

    ⇒ P(E) = 4/50 = 2/25 = 0.08

    Quantity II:

    Total number of cards: 50

    Number of favorable cards = 14 (i.e. 3, 5, 7, 11, 13,17,19,23, 29, 31, 37, 41, 43, 47)

    ⇒ P(E) = 14/50 = 7/25 = 0.28

    ∴ Quantity I < Quantity II

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Nishi Murty
Nishi Murty
Asked: 3 years ago2022-11-08T01:33:37+05:30 2022-11-08T01:33:37+05:30In: Probability

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 II
2. Quantity I ≥ Quantity II
3. Quantity I ˃ Quantity II
4. Quantity I ˂ Quantity II
5. 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 II
2. Quantity I ≥ Quantity II
3. Quantity I ˃ Quantity II
4. Quantity I ˂ Quantity II
5. Quantity I = Quantity II
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  1. ed5a2
    2022-11-11T14:01:08+05:30Added an answer about 3 years ago

    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

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