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2. Draws from a box
Balls, cards, or defective cameras. Two draws, with or without putting the first one back.
| First, then second | Work | Probability |
|---|---|---|
| Green, Green | 4/12 × 3/11 | 0.0909 |
| Green, White | 4/12 × 8/11 | 0.2424 |
| White, Green | 8/12 × 4/11 | 0.2424 |
| White, White | 8/12 × 7/11 | 0.4242 |
| Total | 1 |
- P(Green first)1/3 = 0.33334 of 12
- P(Green then Green)1/11 = 0.09094/12 × 3/11. One Green is gone, so both counts drop by 1.
- P(Green then White)8/33 = 0.24244/12 × 8/11
- P(White then Green)8/33 = 0.24248/12 × 4/11
- P(White then White)14/33 = 0.42428/12 × 7/11
- One of each, either order16/33 = 0.4848Both orders: 8/33 + 8/33. Trap: order matters in the line above, not here.
- At least one Green19/33 = 0.57581 − P(White, White)
- At least one White10/11 = 0.90911 − P(Green, Green)
- P(White second | Green first)8/11 = 0.7273Given, so shrink the box: 8 White left of 11.
- P(Green second | Green first)3/11 = 0.27273 left of 11
- P(Green on second), first unknown1/3 = 0.3333Same as the first draw. Trap question.
- P(Green first | Green second)0.2727Bayes: P(Green,Green) ÷ P(Green second) = 0.0909 ÷ 0.3333
- Are the draws independent?NoP(Green second | Green first) = 0.2727, not 0.3333. Without replacement, the first draw changes the box.