shetin

Anyone who can help me with letters C and D is a godsend. I'm pretty sure I got A and B correct!

Three people get into an elevator on the main floor of a building that has 6 floors (this building has 5 floors above the ground-level). Assuming that each of the three people will get off the elevator on one of floors 2 through 6, and each does so randomly.

Part (a) What is the probability that the three people get off on the 4th floor?(use four decimals) (I got 0.008 )

Part (b) What is the probability that the three people get off on the same floor? (use four decimals) (I got 0.04)

Part (c) What is that at least two of the three people get off on the same floor? (use four decimals)

Part (d) What is the probability that exactly two of the three people get off on the same floor?

Country Boy

[QUOTE=shetin;594852]Anyone who can help me with letters C and D is a godsend. I'm pretty sure I got A and B correct!

Three people get into an elevator on the main floor of a building that has 6 floors (this building has 5 floors above the ground-level). Assuming that each of the three people will get off the elevator on one of floors 2 through 6, and each does so randomly. [quote]
There are 5 floors on which they can get off (2, 3, 4, 5, and 6) so the probability each person gets of on a given one of those is 1/5= 0.20

Yes, the probability is 0.20^3= 0.008.

Yes, 5(0.008 )= 0.04.

"At least two" is "two or three". We have already calculated that the probability three get off on the same floor is 0.04. The probability that two people get off on the same floor, the third on a different floor, is . The probability of "at least two get off on the same floor" is 0.04+ 0.48= 0.52.

The probability is, as calculated above, 0.48.

JeffM1

Country Boy's answer may leave you scratching your head about where 0.48 came from. One way to get there is this:

Person A gets off on floor X with probability 0.2. The probability that person B gets off on floor X given that person A has done so is 0.2 because these are, by hypothesis, assumed to be independent events. The probability that person C gets off on any floor but X given that A and B have got off on 0.8. So the probability that A and B get off on X and C does not is:

$0.2 * 0.2 * 0.8 = 0.032.$

But we are not interested in just floor X, but rather any floor. How many ways can we pick one floor out of five? Nor are we interested in just pair A and B. How many ways can we pick two people out of three?

$\dbinom{5}{1} * \dbinom{3}{2} * 0.032 = \dfrac{5!}{1! * (5 - 1)!} * \dfrac{3!}{2! *(3-2)!} * 0.032 = 5 * 3 * 0.032 = 0.48.$

Another way is this, which seems to be how Country Boy thought about it.

What is the probability that A got off on a floor? By hypothesis, that is 1. What is the probability B got off on that same floor? Obviously 0.2. What is the probability that C got off on a different floor? 0.8. So the probability that A and B get off on the same floor and C gets off on a different floor is

$1 * 0.2 * 0.8 = 0.16.$

But we are not interested just in pair AB, but any pair. How many ways can we pick 2 from 3?

$\dbinom{3}{2} * 0.16 = 3 * 0.16 = 0.48.$