Cards bearing numbers 1, 3, 5, ...., 35 are kept in a bag. A card is d

1.	Cards bearing numbers 1, 3, 5, ...., 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing :(i). A prime number less than 15. (ii). A number divisible by 3 and 5.
Answer» A bag contains cards and each and bearing numbers 1, 3, 5, ……., 35. A cards is drawn at random from the bag. (i). 1, 3, 5, 7, 9, 11 and 13 are prime numbers in the sequence 1, 3, 5, …., 35 which are less than 15. Total prime number less than 15 in given sequence = 7. ∵ Last term is 35 in the given sequence. ∴ 35 = 1+ (n – 1)2 ⇒ 2(n–1) = 35 – 1 = 34 ⇒ n –1 = 17 ⇒ n = 17 + 1 = 18. ∴ Total number of cards = 18. The probability of getting a prime member less than 15 = \(\frac{Total\, prime\, number\, less\, than\, 15\, in \, given\,sequence}{Total\, cards\, in \, the\, bag} = \frac{7}{18}\) (ii). Numbers divisible by 3 and 5 are multiples of 15. Therefore, only 15 are such number in the sequence 1, 3, 5, ……., 35 which are divisible by both 3 and 5. Hence, total number divisible by 3 and 5 in the sequence = 1. (∵15, 30, 45 are multiple of 15 but last term of sequence is 35 & 30 is not odd number) The probability of getting a number divisible by 3 and 5 = \(\frac{Total\, number\, divisible\, by\, 3 and\, 5\, in\, given\, sequence}{Total\, cards\, in\, the\,bag} = \frac{1}{18}\).

Cards bearing numbers 1, 3, 5, ...., 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing :(i). A prime number less than 15. (ii). A number divisible by 3 and 5.

Answer»

A bag contains cards and each and bearing numbers 1, 3, 5, ……., 35. A cards is drawn at random from the bag.

(i). 1, 3, 5, 7, 9, 11 and 13 are prime numbers in the sequence 1, 3, 5, …., 35 which are less than 15.

Total prime number less than 15 in given sequence = 7.

∵ Last term is 35 in the given sequence.

∴ 35 = 1+ (n – 1)2

⇒ 2(n–1) = 35 – 1 = 34

⇒ n –1 = 17

⇒ n = 17 + 1 = 18.

∴ Total number of cards = 18.

The probability of getting a prime member less than 15

= \(\frac{Total\, prime\, number\, less\, than\, 15\, in \, given\,sequence}{Total\, cards\, in \, the\, bag} = \frac{7}{18}\)

(ii). Numbers divisible by 3 and 5 are multiples of 15.

Therefore, only 15 are such number in the sequence 1, 3, 5, ……., 35 which are divisible by both 3 and 5.

Hence, total number divisible by 3 and 5 in the sequence = 1.

(∵15, 30, 45 are multiple of 15 but last term of sequence is 35 & 30 is not odd number)

The probability of getting a number divisible by 3 and 5

= \(\frac{Total\, number\, divisible\, by\, 3 and\, 5\, in\, given\, sequence}{Total\, cards\, in\, the\,bag} = \frac{1}{18}\).