Question

A box has four dice in it. Three of them are fair dice but the fourth one has the number five on all of its faces. A die is chosen at random from the box and is rolled three times and shows up the face five on all the three occasions. The chance that the die chosen was a rigged die, is

• A. $\frac{216}{217}$
• B. $\frac{215}{219}$
• C. $\frac{216}{219}$
• D. none

Answer 1

The correct option is C $\frac{216}{219}$

Given, $A$ is the event which selects the rigged one and $B$ is the event which selects the fair one.
Let E is the event which shows 5 in all three times
Probability of event A for given event E, $P\left(A/E\right)=\frac{1}{4}\ast (1{)}^3=\frac{1}{4}$ ($\frac{1}{4}$ in the equation is probability of selecting one dice among 4)
Probability of event B for given event E, $P\left(B/E\right)=\frac{3}{4}\ast (\frac{1}{6}{)}^3=\frac{1}{1152}$ (since probability of getting 5 in fair dice case=$\frac{1}{5}$)
By Baye's theorem,Probability of selecting the rigged case among both=$\frac{P\left(A/E\right)}{P\left(A/E\right)+P\left(B/E\right)}=\frac{\left(\frac{1}{4}\right)}{\left(\frac{1}{4}\right)+\left(\frac{1}{1152}\right)}=\frac{216}{219}$