Question

Cards marked with numbers 40 , 41 , 42..........60 are well shuffled and a card is drawn at random. What is the probability that the number of the cards is

(i) a prime number

(ii) divisible by 3

(iii) a perfect square

• A. (i) $\frac{1}{4}$ (ii) $\frac{7}{20}$ (iii) $\frac{1}{20}$
• B. (i) $\frac{1}{5}$ (ii) $\frac{3}{20}$ (iii) $\frac{1}{10}$
• C. (i) $\frac{3}{4}$ (ii) $\frac{9}{20}$ (iii) $\frac{1}{10}$
• D. (i) $\frac{1}{2}$ (ii) $\frac{11}{20}$ (iii) $\frac{3}{20}$

Answer 1

The correct option is A

(i) $\frac{1}{4}$ (ii) $\frac{7}{20}$ (iii) $\frac{1}{20}$

Total number of outcome = 20

(i) a prime number

The prime numbers between 40 and 60 are 41 , 43,47,53 and 59

Hence number of favourable outcome = 5

Total outcome = 20

Hence P( a prime number) = $\frac{5}{20}$

= $\frac{1}{4}$

(ii) divisible by 3

Numbers which are divisible by 3 between 40 and 60 are 42,45,48,51,54,57,60

Hence the total number of favourable outcome = 7

Number of possible outcome = 20

Hence P( Number divisible by 3) = $\frac{7}{20}$

(iii) A perfect square

Numbers which are a perfect square between 40 and 60 is only 49

Hence total number of favourable outcome = 1

Number of possible outcome = 20

Hence P( a perfect square ) = $\frac{1}{20}$