an unbiased die is thrown what is the probability of getting an even n

1.	an unbiased die is thrown what is the probability of getting an even number and multiple of 3 2nd a number between 3 and 6
Answer» Answer: In a single throw of an unbiased dice, you can get any ONE of the OUTCOMES: 1, 2, 3, 4, 5, or 6. So, exhaustive number of cases = 6. (i) An even number is OBTAINED if you obtain any one of 2, 4, 6 as an outcome. So, favourable number of cases = 3. Thus, required probability = 3/6 = 1/2 (ii) A multiple of 3 is obtained if you obtain any one of 3, 6 as an outcome. So, favourable number of cases = 2 Thus, required probability = 2/6 = 1/3 (iii) An even number or a multiple of 3 is obtained in any of the following outcomes 2, 3, 4, 6. So, favourable number of cases = 4. Thus, required probability = 4/6 = 2/3 (iv) An even number and a multiple of 3 is obtained if you get 6 as an outcome. So, favourable number of cases = 1. Thus, required probability = 1/6

an unbiased die is thrown what is the probability of getting an even number and multiple of 3 2nd a number between 3 and 6

Answer»

Answer:

In a single throw of an unbiased dice, you can get any ONE of the OUTCOMES: 1, 2, 3, 4, 5, or 6.

So, exhaustive number of cases = 6.

(i) An even number is OBTAINED if you obtain any one of 2, 4, 6 as an outcome.

So, favourable number of cases = 3.

Thus, required probability = 3/6 = 1/2

(ii) A multiple of 3 is obtained if you obtain any one of 3, 6 as an outcome.

So, favourable number of cases = 2

Thus, required probability = 2/6 = 1/3

(iii) An even number or a multiple of 3 is obtained in any of the following outcomes 2, 3, 4, 6.

So, favourable number of cases = 4.

Thus, required probability = 4/6 = 2/3

(iv) An even number and a multiple of 3 is obtained if you get 6 as an outcome.

So, favourable number of cases = 1.

Thus, required probability = 1/6