Discussion

Explanation:

The number of ways of selecting one card from Ashok's bag and other from Shilpa’s bag = 40C1 × 5C1 = 200 ways

Favourable cases:

Cards 1 or 2 or 3: If card 1 or 2 or 3 is drawn from Shilpa’s bad remainder in each case will be less than or equal to 2.

∴ Favourable cases = 40C1 × 3C1 = 120 ways

Card 4: If card 4 is drawn from Shilpa’s bag we need to eliminate those cases when the remainder will be 3.

⇒ We need to eliminate those cards from Ashok’s bag which have number of the form 4x + 3.

∴ Cards of the form 4x + 3 are 3, 7, 11, …., 39 i.e., 10 cards.

∴ Favourable cases = (40 – 10) × 1 = 30 ways

Card 5: If card 5 is drawn from Shilpa’s bag we need to eliminate those cases when the remainder will be 3 or 4.

⇒ We need to eliminate those cards from Ashok’s bag which have number of the form 5x + 3 or 5x + 4.

∴ Cards of the form 5x + 3 are 3, 8, 13, …., 38 i.e., 8 cards.
∴ Cards of the form 5x + 4 are 4, 9, 14, …., 39 i.e., 8 cards.

∴ Favourable cases = (40 - 16) × 1 = 24 ways

∴ Total Favourable cases = 120 + 30 + 24 = 174 ways

⇒ Probability of remainder not greater than 2 = 174/200 = 0.87

Hence, option (b).

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