1.

Tickets are numbered from 10 to 60. A ticket is drawn at random. Find the probability that number drawn on the card is (i) a ticket numbered from 22 to 38 (ii) a prime number (iii) divisible by 3 (iv) divisible by 3 and 2 both .

Answer» Some students take the number of tickets= 60 - 10 = 50 which is wrong.
`" Actually, total number of tickets "= 51" (from 10 to 60)"`
Similarly number of tickets from 22 to 38 is not 16. They are 17 in numbers.
(i) `"Number of favourable cases = 17 (from 22 to 38)"`
`"and total number of cases = 51 (from 10 to 60)"`
`therefore" Required probability ="(17)/(51)=(1)/(3)`
(ii) Prime number from 10 to 60 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53 and 59, i.e., 13 in numbers.
So, required probability `=(13)/(51)`
(iii) Numbers from 10 to 60, which are divisible by 3 are 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60 are 17 in numbers.
So, required probability `=(17)/(51)=(1)/(3)`
(iv) Numbers divisible by 3 and 2 are 6, 12 only
`therefore "Required probability "=(2)/(17)`


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