Question

a box contains 100 cards marked from one to hundred if one card is drawn at random from the box find the probability that it bears first is a single digit number second a number which is a perfect square third is number which is divisible by 7

Answer 1

Total number of outcomes = 100

(i) P (single digit number) = $\bf\huge\frac{9}{100}$

(ii) P (perfect square) =  $\bf\huge\frac{10}{100}$

= $\bf\huge\frac{1}{10}$

(iii) P (a number which is divisible by 7)

=  $\bf\huge\frac{14}{100}$

=  $\bf\huge\frac{7}{50}$

Answer 2

Answer:

Single digit number cards = {1,2,3,4,5,6,7,8,9}=9

Total cards = 100

So, Probability  of getting single digit number = $\frac{9}{100}$

Perfect square number cards = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100} =10

So, probability of getting  number which is a perfect square = $\frac{10}{100}=\frac{1}{10}$

Number which is divisible by 7={7,14,21,28,35,42,49,56,63,70,77,84,91,98}=14

So, probability of getting  number which is divisible by 7 = $\frac{14}{100}=\frac{7}{50}$