There are 8 floors in a building excluding ground floor. 4 people get in the lift along with Mr X on ground floor. Given that each person get off the lift on a different floor, what is the probability that Mr X gets off the lift before exactly 3 people?

Not clear with the explanation for the 5th question, can you please explain a bit further? I can solve it by the traditional method but would like to explore more of this symmetry thing for knowledge,hence asking.

Probability of 3 heads = Probability of 3 tails (which is 2 heads).
Probability of 4 heads = Probability of 4 tails (which is 1 head).
Probability of 5 heads = Probability of 5 tails (which is 0 heads).

dear J
for question 4 if we take the best and worst scenario we will have 9 red and 1 blue and 1 red and 9 blue at the end of all transaction. So can’t we have the range from 0.1 to 0.9 which will include all the options. Please correct me if my approach is wrong.

Yes, but then I would have to calculate the probability for each of those cases separately (using Binomial Probability) which would be a chore. Symmetry allows us to see that overall the answer will be 1/2 irrespective!

There are 8 floors in a building excluding ground floor. 4 people get in the lift along with Mr X on ground floor. Given that each person get off the lift on a different floor, what is the probability that Mr X gets off the lift before exactly 3 people?

Diana, that is one of the questions I had included in today’s post ðŸ™‚ Check it out…in a couple of hours…

regards

J

Not clear with the explanation for the 5th question, can you please explain a bit further? I can solve it by the traditional method but would like to explore more of this symmetry thing for knowledge,hence asking.

Probability of 3 heads = Probability of 3 tails (which is 2 heads).

Probability of 4 heads = Probability of 4 tails (which is 1 head).

Probability of 5 heads = Probability of 5 tails (which is 0 heads).

Now think ðŸ™‚

regards

J

dear J

for question 4 if we take the best and worst scenario we will have 9 red and 1 blue and 1 red and 9 blue at the end of all transaction. So can’t we have the range from 0.1 to 0.9 which will include all the options. Please correct me if my approach is wrong.

Yes, but then I would have to calculate the probability for each of those cases separately (using Binomial Probability) which would be a chore. Symmetry allows us to see that overall the answer will be 1/2 irrespective!

regards

J