# Probability of picking 3 balls from 3 buckets

I have 3 buckets with billiards balls placed as below

• Bucket 1: Balls 1 to 5
• Bucket 2: Balls 6 to 10
• Bucket 3: Balls 11 to 15

1. What is the probability of picking 3 balls at random with one from each bucket?
2. What is the probability of picking 3 balls at random with one from each bucket, with replacement?

What I tried:

The probability of choosing 1st bucket is 1/3 probability of choosing 2nd bucket is 1/3 probability of choosing 3rd bucket is 1/3

Probability of picking up a ball from 1st bucket = 1/5.

Probability of picking up a ball from 2nd bucket = 1/5

Probability of picking up a ball from 3rd bucket = 1/5

So,

Probability of choosing one from 1st bucket

( probability of choosing 1st bucket) x (probability of choosing one ball from 1st bucket) .

(1/3)* (1/5)= 1/15.

similarly from 2nd bucket

(1/3) * (1/5) = 1/15

similarly from 3rd bucket

(1/3)*(1/5) = 1/15

The final probability of picking 3 balls from 3 buckets

1/15 + 1/15 + 1/15 = 3/15

• What have you tried? Note: I'm not sure the problem is clear. How are you selecting the balls? Is it "without replacement" for part $1$? – lulu Feb 4 at 22:17
• HINT for 1) divide the number of ways to pick one from each bucket by the number of ways to pick 3 from all 15 – Bram28 Feb 4 at 22:19
• I pick 3 balls at random no selection. And yes the first one is without replacement – Rob Feb 4 at 22:19
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• To amplify on the previous comment: if we put all 15 balls in one bucket, I could shake the bucket vigorously, reach in without looking and stir the balls around, then (still without looking) grasp whichever ball I can and draw it from the bucket. It is reasonable to suppose the each ball has an equal chance of being chosen, and if I reach in a second and third time, each ball in the bucket at that time has an equal chance to be chosen. But if I have three buckets, what's the procedure? I can only reach into one bucket at a time. – David K Feb 5 at 4:58