Question

30 students enter classroom 319 through two doors. Assume that the probability of entering through the left door is the same as the probability of entering through the right door and is 1/2. What is the probability of the event that 30% of the students enter through the left door and 70% of the students enter through the right door?

Answer 1

This is simply problem of binomial distribution there is two 2 outputs i.e. left door and right door

given that ,

n=30

p= 0.5

let x= number of students enter from left door

the event that 30% of the students enter through the left door = 30*30/100 =9

we want ,the probability of the event that 30% of the students enter through the left door

P( X = 9) = ( 30 choose 9) * ( 0.5 ^9) *(1-0.5)^21

=  0.01332457

The probability of the event that 30% of the students enter through the left door =  0.01332457

Now , 70% of the students enter through the right door

x= 30*70/100 =21

P( X = 21) = ( 30 choose 21) * ( 0.5 ^21) *(1-0.5)^9

P( X = 21) =  0.01332457

The probability of the event that 70% of the students enter through the right door =  0.01332457

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above result give same probability for both event because binomial distribution is symmetric at p =0.5