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
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