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Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Statistics/Probability question help.

I know that (A|B) = P(AnB)/P(B)

But I'm confused here slightly

Prisoner A, Prisoner B and Prisoner C are in seperate cells. They tell prisoner A two/three prisons will be executed and the third will be set free, but prisoner A can't learn of his own fate. Prisoner A estimates therefore that the probability he'll be set free is 1/3

Prisoner A tells the jailer 'I know at least one of the other two prisoners is bound to be executed, so you will
not give anything away if you tell me one of the prisoners to be executed'.

After a little
thought the jailer says `All right, B is to be executed'. Prisoner A now re-estimates his probability of being
set free as 1/2
, since either C or A will be set free!
To analyse this situation, suppose that, in the case when B and C are to be executed, and A is to be set free,
the probability that the jailer says B is to be executed is p. In the other two cases the jailer has no choice.

(1)

Show that the probability that A is to be set free, if we take that that the jailer has said `B is to be executed', equals

p/
(1 + p)
.
(2)

Think about the value of this expression as p takes the values 0, 1
2
and 1 and comment on what the results are.

Just post your question on tutorteddy and get the answer, may be for free.

this is a pretty bad adaptation of the monty hall problem, read about it on wiki : http://en.wikipedia.org/wiki/Monty_Hall_problem

OP

I know, but could you help?

Let

A

be the event that A is set free,

B

be the event that B is set free,

C

be the event that C is set free and

J

be the event that the jailer said "B". From the question,

P(J|A)=p

. Then,

P(A|J)= \frac{P(J|A)P(A)}{P(J)}= \frac{P(J|A)P(A)}{P(J|A)P(A)+P(J|B)P(B)+P(J|C)P(C)}= \frac{p \times \frac{1}{3}}{p \times \frac{1}{3} + 0 \times \frac{1}{3} + 1 \times \frac{1}{3}}= \frac{p}{1+p}

.

OP

Where do you get the P(J|A)P(A) from

and how does (PJ) become so ridiculous?

OP

Original post by answer me now

Let

A

be the event that A is set free,

B

be the event that B is set free,

C

be the event that C is set free and

J

be the event that the jailer said "B". From the question,

P(J|A)=p

. Then,

P(A|J)= \frac{P(J|A)P(A)}{P(J)}= \frac{P(J|A)P(A)}{P(J|A)P(A)+P(J|B)P(B)+P(J|C)P(C)}= \frac{p \times \frac{1}{3}}{p \times \frac{1}{3} + 0 \times \frac{1}{3} + 1 \times \frac{1}{3}}= \frac{p}{1+p}

.

Where does all the stuff on the denominator come from?

Original post by Kolasinac138

Where does all the stuff on the denominator come from?

I'm not telling you because you have no manners.

OP

Ok, whatever

Original post by Kolasinac138

Where does all the stuff on the denominator come from?

It come from the Law of Total Probability: http://en.wikipedia.org/wiki/Law_of_total_probability

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