Here are three sets of cards:

Set A
1, 2, 3, 3, 6, 6, 6, 8, 8, 8

Set B
1, 1, 2, 4, 7, 7, 8, 8, 10, 10

Set C
3, 3, 3, 6, 6, 7, 8, 8, 9

In a game, a player has two options:

Option 1:
Pick two cards from Set A.

Option 2:
Pick one card from Set B and pick one card from Set C.

The cards are picked at random. The player wins if the total of their two cards is exactly 12.

Which option gives a better chance of winning?

Option 1 [ ]
Option 2 [ ]

Show working to support your answer.



Answer :

To determine which option gives a better chance of winning, we need to find the probability of winning for both options and compare them.

### Option 1: Pick two cards from Set A

Set A: 1, 2, 3, 3, 6, 6, 6, 8, 8, 8

1. Total number of ways to choose 2 cards from Set A:
Using combinations, the total number of ways to pick 2 cards from the 10 cards in Set A is:
[tex]\[ \binom{10}{2} = \frac{10!}{2!(10-2)!} = 45 \][/tex]

2. Number of winning combinations (summing to 12):
By examining all possible pairs of cards:
- [tex]\(6 + 6 = 12\)[/tex]
- [tex]\(4 + 8 = 12\)[/tex]

The pairs (indices) from Set A that sum up to 12 are:
- (6, 6)
- (4, 8)

3. Number of winning combinations:
There are 3 valid pairs among the 45 possible pairs that sum exactly to 12.

4. Probability of winning for Option 1:
[tex]\[ P(\text{Win in Option 1}) = \frac{\text{Number of winning pairs}}{\text{Total pairs}} = \frac{3}{45} \approx 0.0667 \][/tex]


### Option 2: Pick one card from Set B and one card from Set C

Set B: 1, 1, 2, 4, 7, 7, 8, 8, 10, 10

Set C: 3, 3, 3, 6, 6, 7, 8, 8, 9

1. Total number of ways to choose one card from Set B and one from Set C:
The total number of combinations is:
[tex]\[ 10 \times 9 = 90 \][/tex]

2. Number of winning combinations (summing to 12):
By examining each card from Set B with each card from Set C:
- [tex]\(4 + 8 = 12\)[/tex]
- [tex]\(7 + 5 = 12\)[/tex]

The pairs (indices) from Set B and Set C that sum up to 12 are:
- (4, 8)
- (7, 5)

3. Number of winning combinations:
There are 2 valid pairs among the 90 possible pairs that sum exactly to 12.

4. Probability of winning for Option 2:
[tex]\[ P(\text{Win in Option 2}) = \frac{\text{Number of winning pairs}}{\text{Total pairs}} = \frac{2}{90} \approx 0.0222 \][/tex]


### Comparison of Probabilities:
- Probability of winning in Option 1: [tex]\(0.0667\)[/tex]
- Probability of winning in Option 2: [tex]\(0.0222\)[/tex]

Option 1 offers a better chance of winning as the probability [tex]\(0.0667\)[/tex] is higher than the probability [tex]\(0.0222\)[/tex] in Option 2.

Thus, the player should choose Option 1.

Option 1 [tex]\(\blacksquare\)[/tex] Option 2 [tex]\(\square\)[/tex]