Proability dice question

eoins23456

Can anybody explain part b?Thanks in advance.

In a bag there are 9 fair dice and one unfair die which never lands on six. An individual picks a die at random from the bag and throws the chosen die k times.Each time he states to an observer whether the die landed on 6 or not.At the end of this procedure the observer must decide whether a fair die or the unfair die was chosen.Let A be the event that the chosen die is fair and B be the event that the die never lands on six.

a) Calculate the probability of even B occuring
Well the probability of getting a six is 9/10* 1/6=3/20
So probability of not getting a six is 17/20
Probability of it never landing on a six is (17/20)*k


b)Suppose B occurs.For what values of K should the observer state that the chosen die is fair,in order to maximise the probability of making a correct decision?
No idea about this one,any suggestions

hivizman

Your analysis of the first part of the problem is, I think, not quite correct. We are told that the individual picks one of the ten dice and then throws that die k times. The way you have calculated the probability assumes that the individual chooses one of the ten dice, throws it once, and then replaces the die, making a fresh random choice of dice each time.

What I think you need to do is work out the probability of throwing no sixes in k throws given either (a) the die chosen is a fair one, or (b) the die chosen is the unfair one. For (b), the probability must be 1, as it's impossible to throw a six with the unfair die. For (a), the probability of not throwing a six in a single throw is 5/6, so the probability of not throwing a six in k throws is (5/6)^k. The probability of choosing a fair die in the first place is 9/10, so overall the probability of not throwing a six is:

(9/10)x(5/6)^k + 1/10

Not that, for k = 1, this gives the same answer as before (17/20), but the probability approaches zero more slowly than the original formula. For example, for k = 2, the original formula gives (17/20)^2 = 0.7225, whereas the revised formula gives a probability of 0.725.

For the second part, we should call "fair" if the probability that the die is fair given that we have thrown no sixes is greater than the probability that the die is unfair given that we have thrown no sixes. This is equivalent to those values of k where (9/10)x(5/6)^k > 1/10, or with a bit of rearrangement, (5/6)^k > 1/9.

The inequality holds for values of k in the range

1 to 12

.