Question

A box contains 90 balls which are numbered from 1 to 90 if one ball is drawn at random from 1 to 90 if one ball is draw at random from the box find the probability that it bears a two digit number a perfect square number a number divisible by 5

Answer 1

Answer:

(i) 9/10 (ii) 1/10 (iii) 1/5

Step-by-step explanation:

total number 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,......, 90

since the common difference between the consecutive number is same

it forms an A.P

first number, a = 10

common difference, d = 11-10 = 1

last number, an = 90

an= a+(n-1)d

90 = 10+(n-1)1

n-1 = 90-10

n = 80 + 1

n= 81

then the favourable number of outcomes = 81

p(two digit number) = p(E) = 81/90

=9/10

(ii) let E be the event of getting 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