Question

0. A die is thrown once. Find the probability of getting (i) a prime number (ii) a number lying between 2 and 6 (iii) an odd number ​

Answer 1

Answer:

Prime numbers = 2, 3 and 5

Favourable number of events = 3

Probability that it will be a prime number =

$\frac{3}{6} = \frac{1}{2}$

Numbers lying between 2 and 6 = 3, 4 and 5 Favourable number of events = 3

Probability that a number between 2 and 6 =

$\frac{3}{6} = \frac{1}{2}$

Odd numbers = 1, 3 and 5

Favourable number of events = 3

Probability that it will be an odd number =

$\frac{3}{6} = \frac{1}{2}$

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Answer 2

Answer:

We use the basic formula of probability to solve the problem.

Number of outcomes on throwing a die is (1, 2, 3, 4, 5, 6) = 6

Number of prime numbers on dice are 1, 3 and 5 = 3

(i) Probability of getting a prime number = Number of prime numbers/total number of outcomes

= 3/6 = 1/2

(ii) Numbers lying between 2 and 6 are 3, 4, 5 = 3

Probability of getting a number lying between 2 and 6 = Number lying between 2 and 6/total number of outcomes

= 3/6 = 1/2

(iii) Total number of odd numbers are 1, 3 and 5 = 3

Probability of getting a odd number = Number of odd numbers/total number of outcomes

= 3/6 = 1/2

Step-by-step explanation:

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