Question

3. 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.

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Answer 1

Explanation:

Total possible events when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of prime numbers = 3 (2, 3 and 5)

➡P (getting a prime number) = 3/6 = ½ = 0.5

(ii) Total numbers lying between 2 and 6 = 3 (3, 4 and 5)

➡P (getting a number between 2 and 6) = 3/6 = ½ = 0.5

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

➡P (getting an odd number) = 3/6 = ½ = 0.5

Answer 2

$\huge\sf \purple{ANSWER:-}$

A die has 6 faces so we have 6 probabilities.

1. prime number from 1 to 6 are :- 2,3,5

so,

$\frac{2}{6} \: and \: \frac{3}{6} \: and \: \frac{5}{6}$

2. 3 and 4 are the numbers between 2 and 6

$\frac{3}{6} \: and \: \frac{4}{6}$

3. let's take an odd number:- 5

$\frac{5} {6}$

Hope it's helpful to uhh......