Probabilty isn't my strong point, but I'll take a run at this one.
So, basically, we first want to count the sets of people who might die - what a morbid problem!!
The total number of people who could die is given by choosing some group of 4 from the 17 = C(17,4) = 2380
Now, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(13,3) = 858 sets
Now, let's look at the possible sets where 2 skiers die - C(3,2) *C(13,2) = 234 sets
Now let's consider the sets where all three skiers die = C(3,3) * C (13,1) = 13
So, the total number of sets containing any of the skiers = 1105
Thus, the chances that all three survive are given by:
1 - (the number of sets containing any of the skiers)/(the total number of possible sets)
= 1 - (858 + 234 + 13)/2380 ≈ 53.6 %
OOPS !! Melody pointed out a math error I made....let me correct this...and as she indicated recently, she DOES believe in "do-overs"....I still like my logic, though!
Now, let's look at the possible sets where 1 skier dies. This is given by C(3,1)*C(14,3) = 1092 sets
Now, let's look at the possible sets where 2 skiers die - C(3,2) *C(14,2) = 273 sets
Now let's consider the sets where all three skiers die = C(3,3) * C (14,1) = 14
So, the total number of sets containing any of the skiers = 1379
Thus, the chances that all three survive are given by:
1 - (the number of sets containing any of the skiers)/(the total number of possible sets)
== 1 - (1379)/2380 ≈ 42.1 %
Thanx, Melody!!
