Home > Homework Forum

# probability qn

• 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) = 7/60 can tell me how to do this question? thanksss
• 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?

my working was
(1/5)(1/4) + (1/3)(1/5) = 7/60

can tell me how to do this question?
thanksss

I'm a Chemistry expert, not a Mathematics expert like Eagle or WeeWS, but this appears to be an extension of the classic (and still excellent) probability riddle known as the "Monty Hall Problem", in which the majority of the population (even some mathematics teachers and professors) gave the wrong answer.

https://en.wikipedia.org/wiki/Monty_Hall_problem

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 :)