One question decides everything: does the first draw change the second?
Two cards are drawn from a deck
Find P(both aces), drawing with replacement.
With replacement the deck is restored, so the draws are independent: (452)(452) = (113)² = 1169. D is the without-replacement answer (452 × 351), which is smaller because one ace has already gone.
the card goes back
the deck resets
the card stays out
the deck has changed
Either way you multiply. What changes is the second probability.
Coin tosses and dice are always independent — a coin has no memory of the last throw. Drawing from a bag or a deck usually is not.
first ace: 452
four aces
second ace: 452
the deck is whole again
P = (452)² = 162704
multiply
= 1169
about 0.6%
first ace: 452
unchanged
second ace: 3/51
one ace and one card gone
P = (452)(351) = 122652
multiply
= 1221
about 0.45%
Writing 452 × 451 is the classic half-correction. If an ace has gone, only three remain — the numerator drops too.
A third draw just adds a third factor, and both numbers drop again: three aces in a row without replacement is 452 × 351 × 250 = 24132 600 = 15525.
| Wording | Which kind |
|---|---|
| “with replacement” | independent |
| “without replacement” | dependent |
| “two cards are drawn at the same time” | dependent |
| “two marbles are taken” | dependent |
| “a coin is tossed twice” | independent |
| “a die is rolled, then a coin flipped” | independent |
Silence usually means without replacement. Taking two objects at once cannot involve putting one back.
A bag of 5 red and 3 blue. Two drawn, no replacement. P(red then blue)?
red first: 58
8 marbles
blue second: 37
blues untouched, total down one
P = 1556
≈ 27%
A red was removed, so the blue count is still 3 — but the total is 7. Which number changes depends on what was taken out, so track it deliberately.
If the question says “one red and one blue in any order”, double it: 1528.
neither: 0.7 × 0.4 = 0.28
both fail: multiply the misses
at least one: 1 − 0.28 = 0.72
the opposite of neither
exactly one: 0.3 × 0.4 + 0.7 × 0.6
A without B, or B without A
= 0.12 + 0.42 = 0.54
two cases, added
Check: both is 0.3 × 0.6 = 0.18, and 0.18 + 0.54 + 0.28 = 1.00 ✓.
That gives 0.1 for “neither”, because it takes away the both-happen part twice. Multiply the two failures instead.
The first success comes late. A player scores with probability 0.4 on each try, independently. What is the probability that the first score comes on the 3rd try?
miss, miss, score
the order is fixed
0.6 × 0.6 × 0.4
a miss is 1 − 0.4
= 0.144
0.36 × 0.4
What is different: “first on the 3rd” fixes the order, so there is only one case and no doubling. The misses before it must be written in.
Testing for independence. P(A) = 0.6, P(B) = 0.3 and P(A and B) = 0.18. Are A and B independent?
P(A) × P(B) = 0.6 × 0.3 = 0.18
what independence predicts
0.18 = P(A and B)
compare with the given value
Yes, independent
had it been 0.2, they would not be
What is different: the rule is used as a test, not a tool. Multiply the two separate chances and compare with the joint chance. Used backwards, it finds a missing chance: if A and B are independent with P(A) = 0.5 and P(A and B) = 0.2, then P(B) = 0.2 ÷ 0.5 = 0.4, and P(A or B) = 0.5 + 0.4 − 0.2 = 0.7.
The second draw on its own. A bag has 5 red and 3 blue marbles. Two are drawn without replacement. What is the probability that the second is red?
red, red: 58 × 47 = 2056
first was red
blue, red: 38 × 57 = 1556
first was blue
P = 3556 = 58
the two cases add
What is different: the first draw is unknown, so both of its outcomes become cases. The answer equals the first-draw chance: unseen, the second marble is just as likely to be red.
47 assumes the first marble was red. Nothing says so; when the first result is not given, split into cases.
red then blue: 38 × 57 = 1556
the total drops to 7
blue then red: 58 × 37 = 1556
P = 3056 = 1528
both orders
1556 is red-then-blue. “One of each” does not say which comes first, so both orders count.
These items mix “and” with a twist: draws without replacement, “neither”, “exactly one”, or one person succeeding while another fails. Write each case as a product, then add the cases. Ten-second check: P(both) + P(exactly one) + P(neither) must be 1.
A box holds 6 white and 4 black balls
Two are drawn without replacement. Find P(both white).
The second draw faces a smaller box: 610 × 59 = 3090 = 13. A uses 610 twice, which is the with-replacement answer. Both the numerator and the denominator drop by one after a success.
