Question: Ashok has a bag containing 40 cards, numbered with the integers from 1 to 40. No two cards are numbered with the same integer. Likewise, his sister Shilpa has another bag containing only five cards that are numbered with the integers from 1 to 5, with no integer repeating. Their mother, Latha, randomly draws one card each from Ashok’s and Shilpa’s bags and notes down their respective numbers. If Latha divides the number obtained from Ashok’s bag by the number obtained from Shilpa’s, what is the probability that the remainder will not be greater than 2?
Explanation:
The number of ways of selecting one card from Ashok's bag and other from Shilpa’s bag = 40 C1 × 5 C1 = 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 = 40 C1 × 3 C1 = 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).