So what we DON"T want is the case of "all even"
Let's find the prob of that.
prob of 3 all even = (4/9)(3/8)(2/7) = 1/27
So prob of at least one odd = 1 - 1/27 = 26/27
For this problem, assume the balls in the box are numbered 1 through 9, and that an experiment consists of randomly selecting 3 balls one after another without replacement.
What probability should be assigned to the event that at least one ball has an odd number?
3 answers
It says that answer is wrong :(
the probability to each outcome is 1/504
I tried 5/504 for my question and it came out wrong
the probability to each outcome is 1/504
I tried 5/504 for my question and it came out wrong
It comes to the same thing!
Each individual outcome is 1/504 (=(1/9)*(1/8)*(1/7)).
There are 4!=24 ways of picking the even numbers, so the probability of picking all even is
24/504=1/21
This is the same as Reiny's number of 2*3*4/(9*8*7), there was a transcription error at the end and became 1/27 instead.
Proceeding with Reiny's logic, the probability of picking at least one odd ball is therefore 1-1/21=20/21 (or 480/504).
Each individual outcome is 1/504 (=(1/9)*(1/8)*(1/7)).
There are 4!=24 ways of picking the even numbers, so the probability of picking all even is
24/504=1/21
This is the same as Reiny's number of 2*3*4/(9*8*7), there was a transcription error at the end and became 1/27 instead.
Proceeding with Reiny's logic, the probability of picking at least one odd ball is therefore 1-1/21=20/21 (or 480/504).
1 answer