Probability Question! (1 Viewer)

[Steve Mate]

Okay all you mathematicians, get ready to flex your Muscles. On one of my usual pricrastinating web searches i stumbled across this tricky, to say the least, question! Lets see what you can do!

A Man inherits on his 21st birthday (31 December, 1970), two houses- one in city and one in the country. On the first day after his birthday, i.e., on 1 January, 1971 he moves into the city house. In this house there is a box containing one red and two white balls and in the country house there is a smiliar box containing one red an three white balls. Each day he draws at random a ball from the box in the house where he is, notes its colour, and then returns it to the box: if it is red he moves to spend the next day at the other house (otherwise he stays where he is and awaits the outcome of the next drawing).

(I) Calculate the probability that he is in residence at the city house:
(a) on 3 January 1971;
(b) on the n-th day after his 21st birthday.
(ii) Assuming that he lives to a ripeold age estimate (approximately) the probability that he will die in the city house!


The answer to your questions is yes! This has appeared in the HSC (1973)
Who ever gets this will have earnt my respect!


Steve!

 

[shimmerz_777]

ok.... i think i need a probability tree...

on jan 1 he has a 2/3 chance of being there the next day, and another 2/3 for the one after that so the chance of him being there 2 days in a row is 4/9

if he goes to the other house instead on the 2nd, he then has a 1/4 chance of changin back... im not sure if i should times this by 1/3 though, but anyways ill say the answer is 25/36 or 19/36

 

[SeDaTeD]

Define P(n) = probability of him being in the city on day n and Q(n) likewise for country. Clearly P(n) + Q(n) = 1.
We have P(1) = 1 since he starts at the city house.

For any day n,
P(n) = Prob[being in city on day n-1 and picking white] + Prob[being in country on day n-1 and picking red]
= (2/3)P(n-1) + (1/4)Q(n-1)
= (2/3)P(n-1) + 1/4 - (1/4)P(n-1)
= (5/12)P(n-1) + 1/4
= (5/12)^2P(n-2) + (5/12)(1/4) + 1/4
...
= (5/12)^(n-1)P(1) + (1/4)[(5/12)^(n-2) + ... + (5/2) + 1]
= (5/12)^(n-1) + (1/4)[(1 - (5/12)^(n-1))/(7/12)]
= (5/12)^(n-1) + (3/7)[1 - (5/12)^(n-1)]
--> 3/7 as n --> infinity.

 

[Steve Mate]

That is pretty awsome.

My respect is yours!


Steve