Question

(iii) A box contains 36 cards bearing only one number from 1 to
36 on each. If a card is drawn at random, find the probability
of an event that the card bears (a) a complete square number
(b) a prime number (c) a multiple of 3.
Don't Spam plz Give Correct Answer ​

Answer 1

Answer:

for better explanation refer the figure

(a) 1/6

(b) 1/3

(c) 1/3

Answer 2

Given:

Cards numbered between 1 to 36

To Find:

Probability of getting

1. a complete square number
2. a prime number
3. multiple of 3

Step-by-step explanation:

• Total number of complete square numbers between 1-36 = 1, 4, 9, 16, 25, 36⇒ P(E)=6

        Probability(complete square number)= $\frac{6}{36}$

                                                                     = $\frac{1}{6}$

       ∴ The probability of getting the complete square number is 1/6

•  Total number of prime numbers between 1-36 =     2,3,5,7,11,13,17,19,23,29,31 ⇒ P(E)= 11

         Probability(Prime number)= $\frac{11}{36}$

      ∴ The probability of getting a prime number is 11/36

• Numbers which are multiple of 3 between 1-36 = 3,6,9,12,15,18,21,24,27,30,33,36 ⇒ P(E)= 12
•   Probability(Prime number)= $\frac{12}{36}$
•                                               =$\frac{1}{3}$
• ∴
• The probability of getting a multiple of 3 is 1/3