Puzzle 16 of 26 · Hard

The last colour standing

A bag holds 3 red, 2 blue and 2 green marbles. You take them out one at a time, at random. What’s the chance that when the last red marble comes out, at least one blue and at least one green are still in the bag?

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Answer: 8/35 ≈ 22.9%

Both orange paths leave a blue and a green behind when the last red comes out.

Step 1 · The last marble of all is blue
$\displaystyle 2 \text{ of } 7 = \frac{2}{7}$
Blue and green must both outlast red, so the very last marble is blue or green. Split on which one it is.
Step 2 · Then green outlasts red
$\displaystyle 2 \text{ green of the } 5 \text{ red and green } = \frac{2}{5}$
The reds and greens still come out in a random order among themselves, so the last of them is green with chance 2/5.
Step 3 · Blue last, then green outlasts red
$\displaystyle \frac{2}{7} \times \frac{2}{5} = \frac{4}{35}$
Both things must happen on this path, so multiply. Green last works the same way with the colours swapped.
Step 4 · Add both cases
$\displaystyle \frac{4}{35} + \frac{4}{35} = \frac{8}{35}$
The last marble can’t be both blue and green, so the two cases don’t overlap. Add them.

Answer $\displaystyle \frac{8}{35} \approx 22.9\%$

Step 4 of 4

The method · Unit 3, Conditional probabilitySplitting into casesFind a chance by adding up every case

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