Q. 285.0( 6 Votes )

# A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number, (ii) a perfect square number, (iii) a number divisible by 5.

Answer :

Total numbers of elementary events are: 90

(i) Let E be the event of getting two digit number at a draw

The favourable numbers are: 10, 11, 12, 13, 14 ……… 90

Since the common difference between the consecutive number is same

It forms an A.P.

First number = a = 10

d = common difference = 11- 10 = 1

Last number = a_{n} = 90

a_{n} = a + (n-1) d

90 = 10 + (n-1) 1

90-10 = (n-1)

80 + 1 = n

81 = n, being number of terms

Then, the favourable numbers of outcome = 81

∴ P (two digit number) = P (E) = 81/90 = 9/10

(ii) Let E be the event of getting a perfect square number

The favourable numbers are: 1, 4, 9, 16, 25, 36, 49, 64, 81

Then, the number of favourable outcomes = 9

∴ P (perfect square number) = P (E) = 9/90 = 1/10

(iii) Let E be the event of getting a number divisible by 5

The favourable numbers are: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90

Then, the number of favourable outcomes = 18

∴ P (number divisible by 5) = P (E) = 18/90 = 1/5

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