You have a 8 faced die having numbers from 1 to 8 on its 8 faces. What would be the probability of getting 8 on throwing it?

1/6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1/3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1/8

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1/2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
1/8

If there are 6 faces, the probability of getting one of 6 digits is $\frac{1}{6}$ . So, in the case of 8 faces, probability of getting any of the 8 numbers must be $\frac{1}{8}$ . Hence, the probability of getting 8 on throwing it is $\frac{1}{8}$ .

Suggest Corrections

Similar questions

Q. You have a six-faced dice having numbers from

1

6

on its faces. What will be the probability of getting

5

on throwing it?

The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is

(A) 1 (B) 2 (C) 5 (D)

There is an octagonal die with 8 faces having values 1 to 8 on each face. It is thrown 4 times in a sequence. In how many ways can you get a sum which is less than or equal to 20?

In an event of throwing the dice, what is the probability of getting a number 4, if the numbers on the dice start from 0 and has six faces?

Q. Two dices are rolled. If both dices have six faces numbered

1, 2, 3, 5, 7

and

11,

then the probability that the sum of the numbers on the top faces is less than or equal to

8

is:

Join BYJU'S Learning Program

You have a 8 faced die having numbers from 1 to 8 on its 8 faces. What would be the probability of getting 8 on throwing it?

The correct option is C 1/8 If there are 6 faces, the probability of getting one of 6 digits is 16. So, in the case of 8 faces, probability of getting any of the 8 numbers must be 18. Hence, the probability of getting 8 on throwing it is 18.

The correct option is C
1/8

If there are 6 faces, the probability of getting one of 6 digits is $\frac{1}{6}$ . So, in the case of 8 faces, probability of getting any of the 8 numbers must be $\frac{1}{8}$ . Hence, the probability of getting 8 on throwing it is $\frac{1}{8}$ .