In a block of ten cards (tickets) there are five cards worth…
In a block of ten cards (tickets) there are five cards worth $100$ dinars, three cards worth $300$ dinars and two cards worth $500$ dinars. Three cards are drawn at random. What is the probability that:
$1^{\circ}$ at least two of the drawn cards have the same value;
$2^{\circ}$ the total value of the three drawn cards is $700$ dinars?
Explanation of Method
All groups of $3$ cards out of $10$ are equally likely: $\binom{10}{3}=120$ outcomes. For $1^{\circ}$ the opposite event is simpler: "all three values are different", which means one card of each value. For $2^{\circ}$ we list all ways to write $700$ as a sum of three card values, and count the groups for each way with combinations.
Step 1: All outcomes
\[ \binom{10}{3} = \frac{10\cdot9\cdot8}{3\cdot2\cdot1} = 120. \]
| Value | 100 | 300 | 500 |
|---|---|---|---|
| Number of cards | 5 | 3 | 2 |
Step 2: $1^{\circ}$ — at least two cards with the same value
The opposite event: all three values are different. With only three values available, the three cards must be one $100$, one $300$ and one $500$:
\[ 5\cdot3\cdot2 = 30 \text{ groups}, \qquad P(\text{all different}) = \frac{30}{120} = \frac14. \]
\[ \boxed{P_1 = 1 - \frac14 = \frac34 = 0.75.} \]
Step 3: $2^{\circ}$ — total value 700 dinars
Which three values from $\{100, 300, 500\}$ add up to $700$?
| Combination | Check | Number of groups |
|---|---|---|
| $100+100+500$ | $700$ ✓ | $\binom52\binom21=10\cdot2=20$ |
| $100+300+300$ | $700$ ✓ | $\binom51\binom32=5\cdot3=15$ |
| $100+100+100$ | $300$ | — |
| $100+100+300$ | $500$ | — |
| $100+300+500$ | $900$ | — |
| $300+300+300$ and larger | $\ge900$ | — |
Favourable groups: $20 + 15 = 35$.
\[ \boxed{P_2 = \frac{35}{120} = \frac{7}{24} \approx 0.292.} \]
Step 4: Check part $1^{\circ}$ directly
"At least two equal" = "exactly two equal" or "all three equal":
- all three equal: $\binom53 + \binom33 = 10 + 1 = 11$ (three 500s are impossible);
- exactly two equal: $\binom52\cdot5 + \binom32\cdot7 + \binom22\cdot8 = 50 + 21 + 8 = 79$.
Total $11 + 79 = 90$, and $90/120 = 3/4$. ✓
Result
| Part | Favourable / all | Probability |
|---|---|---|
| $1^{\circ}$ at least two equal values | $120-30=90$ of $120$ | $3/4=0.75$ |
| $2^{\circ}$ total 700 dinars | $20+15=35$ of $120$ | $7/24\approx0.292$ |