From a deck of 36 cards, three cards are drawn at random.…
From a deck of $36$ cards, three cards are drawn at random. Find the probability that among them there will be at least one ace.
Explanation of Method
"At least one ace" is the opposite of "no ace". The probability of no ace is easy: all three cards come from the $32$ non-aces. Subtracting from $1$ gives the answer. As a check we add the probabilities of exactly one, two and three aces.
Step 1: All outcomes
\[ \binom{36}{3} = 7140 \text{ equally likely groups of three cards}. \]
Step 2: The opposite event — no ace
\[ \binom{32}{3} = \frac{32\cdot31\cdot30}{6} = 4960, \qquad P(\text{no ace}) = \frac{4960}{7140} = \frac{248}{357}. \]
Step 3: The answer
\[ \boxed{P(\text{at least one ace}) = 1 - \frac{248}{357} = \frac{109}{357} \approx 0.305.} \]
Step 4: Check by cases
| Number of aces | Groups | Probability |
|---|---|---|
| $0$ | $\binom40\binom{32}{3}=4960$ | $0.6947$ |
| $1$ | $\binom41\binom{32}{2}=1984$ | $0.2779$ |
| $2$ | $\binom42\binom{32}{1}=192$ | $0.0269$ |
| $3$ | $\binom43\binom{32}{0}=4$ | $0.0006$ |
| total | $7140$ | $1$ ✓ |
$1984 + 192 + 4 = 2180$ groups contain at least one ace, and $2180/7140 = 109/357$. ✓ (See Problem 48 for the case of exactly one ace.)
Result
| Answer | |
|---|---|
| $P(\text{no ace})$ | $\binom{32}3/\binom{36}3=248/357$ |
| $P(\text{at least one ace})$ | $109/357\approx0.305$ |