I'm hurting while trying to figure out these questions...




Apparently, those answers are not right. Can someone check them and tell me what I did wrong please?

For the first one, you're working with a binomial distribution.
A person can either show up for their flight, or not show up for their flight.

The probability of showing up is (1 - the probability of not showing up)
Or (1-0.0977) = .9023

There are not enough seats on the plane when either 17 or 18 people decide to show up --> More than 16 people show up.
Using the "binomial formula" you can calculate the probability of more than 16 "successes."

From the looks of it, the probability of more than 16 people showing up is roughly 0.46344.

This post was edited by MidnightRider on Jan 21 2014 12:19am

Probability that 17 or 18 people show up is ~ 46.34 %

( 1 - 0.0977 )^18 + 18* ( 1 - 0.0977 )^17 * 0.0977

I let you decide if it's wise or not to overbook the flight.

A Poisson law with mean L gives you :
P( X = k ) = exp( -L ) * L^k / k!

In your homicide problem, L = 97 / 280
so :

P( X = 0 ) = exp ( - 97/280 ) ~ 0.7072

P( X = 1 ) = exp ( -97/280 ) * (97/280) ~ 0.2450

P( X = 2 ) = exp ( -97/280 ) * (97/280)² / 2 ~ 0.0424

P( X = 3 ) = exp ( -97/280 ) * (97/280)^3 / 6 ~ 0.0049

P( X = 4 ) = exp ( -97/280 ) * (97/280)^4 / 24 ~ 0.0004

Compare this to :
197/280
70/280
12/280
1/280
0/280

Seems to match the results good enough...