Question

A bag contains three dice, two of which are somewhat unusual: die A is a regular die whose faces show 1,2,3,4,5 and 6. Die B has a 4 on all 6 sides, and the faces of die C are2,2,4,4,6and 6. Suppose that a die is randomly selected, rolled, and comes up a 4. If the same die is rerolled, what is the probability that it is again a 4?

Answer 1

Answer:

i dont know

Step-by-step explanation:

Answer 2

Answer:

41/54

Step-by-step explanation:

I will provide the approach and work it out.  However, I am error prone, so check the arithmetic.

This is a Bayes Theorem problem.

Consider 3 "causes for a 4", Die A, B, or C, and an event, E.

The probability of each cause, given the event 4 is

$P(A|4)=\frac{P(A)P(4|A)}{P(A)P(4|A)+P(B)P(4|B)+P(C)P(4|C)}$

$P(B|4)=\frac{P(B)P(4|B)}{P(A)P(4|A)+P(B)P(4|B)+P(C)P(4|C)}$

$P(C|4)=\frac{P(C)P(4|C)}{P(A)P(4|A)+P(B)P(4|B)+P(C)P(4|C)}$

Since the rolled die is chosen at random,

P(A)=P(B)=P(C)=1/3

Also, from the numbers on the faces of each die,

P(4|A)=1/6

P(4|B)=1

P(4|C)=1/3

Making all the substitutions, we get

P(A|4)=1/9

P(B|4)=6/9

P(C|4)=2/9

That these sum to 1 is promising (maybe I didn't make a mistake).

Next we must roll the die again and the probability of a 4 could be because the die is A, B, or C.  Thus, the desired probability is given by

P(A|4)P(4|A)+p(B|4)P(4|B)+P(C|4)P(4|C)

This evaluates to

(1/9)(1/6)+(6/9)(1)+(2/9)(1/3)=41/54