Question

a student had rolled 114 dice and removed all those landed on the number six. the student rolled the remaining dice again and removed those that landed on a six. when the student had rolled the dice 20 times there were 9 dice left.

Calculate the most likely number of times that the student had rolled the dice before the number of dice had halved.​

Answer 1

Answer:

4 rolls

Explanation:

The probability of rolling a 6 is 1/6 = 0.1667 (4.d.p)

This means that approx. 1/6 of the dice will land on a 6.

If the student removes a 1/6 of the dice each time, it means that the number of dice is decreasing by 1/6 each roll (or multiplying by 5/6).

114/2 = 57

By multiplying 114 by 5/6 and continuing we can see how many times it takes for us to get past 57.

Roll 1: 114 x 5/6 = 95

Roll 2: 95 x 5/6 = 79.17

Roll 3: 79.17 x 5/6 = 65.97

Roll 4: 65.97 x 5/6 = 54.98

Most likely to take 4 rolls before halving the number of dice.

You could also solve this problems by writing an equation and solving using logs but I think this method is easier to understand.

Hope this helped!