Question

A die is thrown. Find the probability of getting
i. 0
ii. a 2 – digit number

Answer 1

ii. a 2-digit number

Step-by-step explanation:

Throwing of a die can have 6 possible outputs.

1, 2, 3, 4, 5 and 6.



So, N(S)N(S) (number of Space) is 6

The numbers having greater value than 22 are- 3, 4, 5 and 6.

So, we can get a number which is greater than 2 in four ways.

So, N(E)N(E) (number of Events) is 4

If probability of finding a number greater than 4 in throw of a die is denoted by P(A)P(A), then-

P(A)=N(E)N(S)=46=23P(A)=N(E)N(S)=46=23 (a)

There you go!

Answer 2

i) Since there are 6 possibilities on a regular die (1, 2, 3, 4, 5, 6), and a 0 is not one of those values stated, there is a 0/6 probability of getting a 0.
ii) Since there are 6 possibilities on a regular die (1, 2, 3, 4, 5, 6), and there are no two digit numbers in those values stated, there is a 0/6 probability of getting a two digit number.