Question

A box contains 90 cards, which are numbered from 1 to 90. If one card is drawn at random from the box, find the probability that it bears. (i) a three digit number (ii) a perfect cube number (iii) a number divisible by 9

Answer 1

Answer:

(1) from 1 to 90 there is not any 3 digit no. , it's an impossible case.So probability of getting 3 digit no.is 0

(2) perfect cube are 1,8,27,64

p(perfect cube)=4/90=0.044

(3)p(no. divisible by 9)=10/90=1/9