Probability • Topic 5 of 5

Dependent Events

Events are dependent when the first outcome changes the probabilities for the second — most commonly drawing items without replacement. After the first draw, both the favorable count and the total shrink by one, so the second probability uses the updated numbers. To find the probability that both of two draws are red from a box of r red out of t total, compute (r/t) × ((r−1)/(t−1)). The defining feature is “without replacement”; if the item is replaced, the events become independent. Track how each draw changes the counts, multiply the adjusted probabilities, and reduce the final fraction.

A probability tree for drawing two marbles without replacement, showing the second probability changing as totals shrinkDependent events6 totalred 4/6blue 2/6now 5 totalred 3/5blue 2/5Without replacement: the second probability changes

✅ Solved examples

1. A box has 4 red, 2 blue. Two drawn without replacement. P(both red)?
(4/6)(3/5) = 12/30 = 2/5.
2. A bag has 3 red, 3 green. Two drawn, no replacement. P(both green)?
(3/6)(2/5) = 6/30 = 1/5.
3. A box has 5 red, 5 blue. P(first red, then blue), no replacement?
(5/10)(5/9) = 25/90 = 5/18.
4. A box has 2 red, 4 blue. Two drawn, no replacement. P(both red)?
(2/6)(1/5) = 2/30 = 1/15.

✏️ Practice — try these, take hints as needed

1. A box has 3 red, 5 blue. Two drawn, no replacement. P(both red)?
(3/8)(2/7).
Multiply.
Reduce.
3/28.
2. A bag has 4 green, 2 red. Two drawn, no replacement. P(both green)?
(4/6)(3/5).
12/30.
Reduce.
2/5.
3. A box has 6 red, 4 blue. Two drawn, no replacement. P(both blue)?
(4/10)(3/9).
12/90.
Reduce.
2/15.
4. A bag has 5 red, 3 blue. Two drawn, no replacement. P(both red)?
(5/8)(4/7).
20/56.
Reduce.
5/14.
5. A box has 3 red, 3 blue. Two drawn, no replacement. P(both red)?
(3/6)(2/5).
6/30.
Reduce.
1/5.

📝 Topic test — 8 questions

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