Question

A cubical die has two faces numbered 1, two numbered 2 and two numbered 3, and is such that when rolled, the probability of scoring 1 is 1/4, scoring 2 is 1/2 and scoring 3 is 1/4.
If the die is rolled 3 times, show that the probability that the smallest score will be 1 and the largest score will be 3 is equal to 9/32.

Answer 1

http://www.wolfram.com/mathematica/new-in-8/probability-and-statistics-solvers-and-properties/

Answer 2

Solver^^

Answer 3

wolfram can solve probability?! since when?

Answer 4

select the tye of ques... and choose the type of prob... and get it dolved

Answer 5

ok thanks!

Answer 6

The probability of throwing a one in three rolls of the die is:\[\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\]
The probability of throwing a tree in three rolls of the die is:\[\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\]
Throwing a two is allowed on only one of the three throws. The probability of not throwing a 2 on two rolls is \[1-\frac{1}{2}=\frac{1}{2}\]
Therefore the required probability is\[\frac{3}{4}\times \frac{3}{4}\times \frac{1}{2}=\frac{9}{32}\]