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 exactly one ace.
Explanation of Method
A 36-card deck has $4$ aces and $32$ other cards. All groups of $3$ cards are equally likely. A favourable group consists of $1$ ace (from $4$) and $2$ non-aces (from $32$). We count with combinations and divide — the hypergeometric formula from Problem 37.
Step 1: The deck
A 36-card deck has $4$ suits with $9$ ranks each (6, 7, 8, 9, 10, J, Q, K, A). So there are $4$ aces and $36-4=32$ other cards.
Step 2: All outcomes
\[ \binom{36}{3} = \frac{36\cdot35\cdot34}{3\cdot2\cdot1} = 7140. \]
Step 3: Favourable outcomes
- choose the ace: $\binom41 = 4$ ways;
- choose the other two cards from the $32$ non-aces: $\binom{32}{2} = \frac{32\cdot31}{2} = 496$ ways.
\[ \text{favourable} = 4\cdot496 = 1984. \]
Step 4: The probability
\[ \boxed{P = \frac{\binom41\binom{32}{2}}{\binom{36}{3}} = \frac{1984}{7140} = \frac{496}{1785} \approx 0.278.} \]
Step 5: Check with the one-by-one method
The ace can be the first, second or third card ($3$ positions). For "ace first":
\[ \frac{4}{36}\cdot\frac{32}{35}\cdot\frac{31}{34}, \qquad P = 3\cdot\frac{4\cdot32\cdot31}{36\cdot35\cdot34} = \frac{11904}{42840} = \frac{496}{1785}. \ \checkmark \]
Result
| Answer | |
|---|---|
| Favourable / all | $\binom41\binom{32}2=1984$ of $\binom{36}3=7140$ |
| Probability | $496/1785\approx0.278$ |