How do you solve this problem?

Question

Assume the balls in a 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?

Answer 1

P(at least one odd) = 1 - P(all even)

P(all even) = 4/9 x 3/8 x 2/7 = 1/27

For the first ball, there are 4 even numbers out of 9 balls. After that is picked, there are 3 even numbers out of 8 balls. Finally the third pick there are 2 even numbers out of 7 balls.

P(at least one odd) = 1 - P(all even)

So the chance there is at least one odd is 1 - 1/27 = 26/27.

This is approximately ≈96.3%

Answer 2

don't worry im on the case! (see middle for explaination, or just scroll down to see the answer)

its a pretty high probability. figure theres 504 possibilities, since you could draw any one of nine balls on the first time, any one of eight balls on the second time, and any one of seven balls on the third time. 9*8*7 is 504. with one number set, for instance once the first ball is drawn, there are 8*7 possibilities, or 56. so for the five times out of nine you draw an odd on the first pick, you can add 56 to the pool of things that work. (n.b-the final probability will be the total things that work/504). so far we have 280. now, for the four out of nine times you get an even on the first draw. i figured i'd just list all the possibilities. say you got a two on the first draw. if you get an odd on the second, (5/8 chance) you can add 7 more to the pool, since no matter WHAT you get on the third time you got your odd. so add 35 more! and then say you get a 2, then a 4. there are still five possibilities left! so because of 2,4; we can add five more to the pool. so theres five more for 2,6; 2, 8. total for drawing two on the first pick: 50 added. however, that will be the same for if you get a 4, or a 6, or an 8 on the first pick, since we just want an odd...0,1, or 2 evens...the specifics don't matter. so we add the 200 possibilities of getting an odd when you draw an even on the first pick to the 280 we had earlier, and we get 480! so the answer is 480/504, but we need to simplify. thats real ez. it becomes 240/251. 251 is prime, so it cant be simplified further.


FINAL ANSWER: 240/251 YAY!!!

Answer 3

So, how do you compute the probability of getting the all three balls with even numbers?

Sample space 1 2 3 4 5 6 7 8 9

Number of odd: 5
Number of even: 4

P(3 balls even) = 4/9 * 3/8 * 2/7

P(1 odd) = 1 - P(3 balls even) = 1-0.04761= 0.95238

Good luck

Answer 4

P = 1 - (# of permutations of three even balls)/(# of permutations of three balls) = 1 - (4!/1!)/(9!/6!) = 1 - 24/504 = 1 - 1/21 = 20/21 ≈ 95.2%

Bandf: your method is valid, but you've made an arithmetical error. 4/9 * 3/8 * 2/7 = 1/21, not 1/27.