Problem 42 · AMC 8 Stretch
Stretch
Counting & Probability
binomial-probabilitycomplementary-counting
A card is drawn from a full deck and put back; this is done 3 times. Find the probability of getting at least one ace in the 3 draws. (An ace has probability \(\tfrac{4}{52} = \tfrac{1}{13}\) each draw.)
Show answer
Answer: 469/2197 (about 0.21)
Show hints
Hint 1 of 4
Because the card is put back, the draws are independent with the same ace-chance \(1/13\) (and no-ace chance \(12/13\)).
Still stuck? Show hint 2 →
Hint 2 of 4
'At least one ace' is awkward to count directly (1, 2, or 3 aces). It is far easier to count the OPPOSITE: no aces at all.
Still stuck? Show hint 3 →
Hint 3 of 4
\(P(\text{no aces in 3 draws}) = (12/13)^3\). Then \(P(\text{at least one}) = 1\) minus that.
Show solution
Approach: Complementary counting — 1 minus 'no aces'
- Putting the card back makes the draws independent. Each draw is NOT an ace with probability \(\frac{12}{13}\).
- Directly counting 'at least one ace' means handling 1, 2, or 3 aces — messy. The slick move is to find the chance of NO aces and subtract from 1.
- \(P(\text{no aces}) = \left(\frac{12}{13}\right)^3 = \frac{1728}{2197}\).
- \(P(\text{at least one ace}) = 1 - \frac{1728}{2197} = \frac{469}{2197} \approx 0.21\).
Mark:
· log in to save