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?
Question
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
[tex]P(A|4)=\frac{P(A)P(4|A)}{P(A)P(4|A)+P(B)P(4|B)+P(C)P(4|C)}[/tex]
[tex]P(B|4)=\frac{P(B)P(4|B)}{P(A)P(4|A)+P(B)P(4|B)+P(C)P(4|C)}[/tex]
[tex]P(C|4)=\frac{P(C)P(4|C)}{P(A)P(4|A)+P(B)P(4|B)+P(C)P(4|C)}[/tex]
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
Answer:
i dont know
Step-by-step explanation: