bmwhype2 wrote:
I'm having trouble with AT LEAST in combinatorics.
we have 4 slots. There are 8 digits to fill those spots: the digits 2 through 9.
What is the probability that those 4 slots will contain AT LEAST one prime number?
What is the probability that those 4 slots will contain AT LEAST TWO prime numbers?
What is the probability that those 4 slots will contain AT LEAST THREE prime numbers?
Assuming no restrictions (i.e. can repeat the digits).
P(At least 1 Prime) : 1 - P(No Prime)
p(no prime) : 2^-4 = 1/16 ---> selecting 4,6,8,or 9 four times
so 1- (1/16) = 15/16
P(At least 2 Prime) : 1 - [P(No Prime) + P(1 prime)]
p(1 prime) = C(4,1) * (1/2)*(1/2)*(1/2)*(1/2) = 4* 2^-4 = 1/4
*note that the prob of selecting primes equals that of selecting non-primes--1/2 or 4/8
so 1 - [(1/16) + (1/4)] = 11/16
P(At least 3 Prime) : 1 - [P(No Prime) + P(1 prime)+ P(2 prime)]
p(2 prime) = C(4,2) * (1/2)*(1/2)*(1/2)*(1/2) = 6* 2^-4 = 3/8
*note that the prob of selecting primes equals that of selecting non-primes--1/2 or 4/8
so 1 - [(1/16) + (1/4) + (3/8)] = 5/16
If there's a shorter way, i'd like to know. I think it gets much more confusing when there are restrictions such as "exactly 2 primes together".
For instance, what is the probability of getting 3 primes, two of which are back to back - e.g.
238
5