Probability • Topic 2 of 4

Independent & Dependent Events

For a sequence of events joined by "and", you multiply — but which numbers you multiply depends on whether the events are independent or dependent. Events are independent when the first does not change the second: coin tosses, dice rolls, or drawing with replacement. Then P(A and B) = P(A) × P(B). Events are dependent when the first alters the sample space for the second — the signature case is drawing without replacement, where the count and total both drop. Then P(A then B) = P(A) × P(B given A). Deciding "with or without replacement" is the whole game: two aces drawn with replacement is (4/52)(4/52), but without replacement it is (4/52)(3/51), because one ace is gone and the deck now has 51 cards. The GRE almost always means "without replacement" when it says the items are not put back — read that clause carefully, because it silently changes every number after the first draw.

✅ Solved examples

1. A fair coin is tossed twice. Probability of heads both times?
Independent tosses: P = 1/2 × 1/2 = 1/4.
2. A bag has 5 red and 3 blue balls. Two are drawn WITHOUT replacement. Probability both are red?
Dependent: first red 5/8, then 4 reds of 7 remain → 4/7. P = 5/8 × 4/7 = 20/56 = 5/14.
3. The same bag (5 red, 3 blue), but two draws WITH replacement. Probability both red?
Independent now: P = 5/8 × 5/8 = 25/64. The ball is returned, so the second draw is unchanged.
4. Quantitative Comparison. From a deck, two cards are drawn. Quantity A: P(both aces) WITH replacement. Quantity B: P(both aces) WITHOUT replacement. Which is greater?
With replacement: (4/52)(4/52) = 16/2704 = 1/169. Without: (4/52)(3/51) = 12/2652 = 1/221. Since 1/169 > 1/221, Quantity A is greater — removing an ace lowers the second draw's odds.

✏️ Practice — try these, take hints as needed

1. A die is rolled twice. Probability of a 6 both times?
Independent rolls.
1/6 × 1/6.
Multiply.
1/36
2. A bag has 4 red and 6 green. Two drawn without replacement — probability both green?
First green: 6/10.
Then 5 greens of 9 remain: 5/9.
6/10 × 5/9.
1/3
3. A coin is tossed 3 times. Probability of three tails?
Independent tosses.
(1/2)³.
Cube it.
1/8
4. From 3 red and 2 blue, two are drawn without replacement. Probability both red?
First red: 3/5.
Then 2 reds of 4 remain: 2/4.
3/5 × 1/2.
3/10
5. A spinner lands on red with probability 1/3 each spin. Probability of red on two independent spins?
Independent spins multiply.
1/3 × 1/3.
Multiply.
1/9

📝 Topic test — 8 questions

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