Originally posted by Jasmine ngjiamin:In a monty hall game, you are given 5 doors. Behind 1 of the 5 doors there is a big prize, and the rest there is nothing. Assume that you have already chosen door 1, and the host has opened one of the doors without the prize (i.e. Door2). What is the probability that the prize is behind door 1 or door 3?
Answer: 7/15
my working was
(1/5)(1/4) + (1/3)(1/5) = 7/60can tell me how to do this question?
thanksss
This should help you
http://math.stackexchange.com/questions/608957/monty-hall-problem-extended
(1/5) + (4/5)*(1/3) = 7/15
1/5 refers to probability of choosing Door 1
4/5 refers to probability of not choosing Door 1, and 1/3 refers to probability of choosing Door 3 out of Door 3, 4 or 5.
Thanks for sharing the question and its answer :)