Answered

Here are three sets of cards.

Set [tex]$A$[/tex]:
1, 2, 3, 3, 6, 6, 6, 8, 8

Set [tex]$B$[/tex]:
1, 1, 2, 4, 7, 7, 8, 8, 10, 10

Set [tex]$C$[/tex]:
3, 3, 3, 6, 6, 7, 8, 8, 9

In a game, a player has two options.

\begin{tabular}{|c|}
\hline
Option 1 \\
Pick two cards from Set [tex]$A$[/tex] \\
\hline
\end{tabular}

Option 2
\begin{tabular}{|c|}
\hline
Pick one card from Set [tex]$B$[/tex] and pick one card from Set [tex]$C$[/tex] \\
\hline
\end{tabular}

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 [tex]$\square$[/tex] Option 2 [tex]$\square$[/tex]

Show working to support your answer. [4 marks]



Answer :

GinnyAnswer
To determine which option gives a better chance of winning, we need to calculate the probability of winning with each option.

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

First, let's identify all the pairs of cards from Set A that sum up to 12. The cards in Set A are:
[tex]\[ 1, 2, 3, 3, 6, 6, 6, 8, 8 \][/tex]

To find pairs that sum to 12, we check each combination:
- [tex]\(1 + 11 = 12\)[/tex] but 11 is not in the set.
- [tex]\(2 + 10 = 12\)[/tex] but 10 is not in the set.
- [tex]\(3 + 9 = 12\)[/tex] but 9 is not in the set.
- [tex]\(6 + 6 = 12\)[/tex] (valid pair)
- [tex]\(8 + 4 = 12\)[/tex] but 4 is not in the set.

The valid winning pairs from Set A are [tex]\((6, 6)\)[/tex], [tex]\((6, 6)\)[/tex], and [tex]\((6, 6)\)[/tex]:

We found 9 pairs:
[tex]\[ (6, 6), (6, 6), (6, 6), (6, 6), (6, 6), (6, 6), (6, 6), (6, 6), (6, 6) \times \text{9 times for every } 6 \][/tex]

Next, we calculate the total number of possible pairs in Set A. Since there are 9 cards in Set A, the total number of pairs is:
[tex]\[ 9 \times 9 = 81 \][/tex]

Now, calculate the probability of winning with Option 1:
[tex]\[ \frac{\text{Number of winning pairs}}{\text{Total number of pairs}} = \frac{9}{81} = 0.1111 \][/tex]

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

The cards in Set B are:
[tex]\[ 1, 1, 2, 4, 7, 7, 8, 8, 10, 10 \][/tex]

The cards in Set C are:
[tex]\[ 3, 3, 3, 6, 6, 7, 8, 8, 9 \][/tex]

To find pairs that sum to 12, we check each combination where one card is from Set B and one card is from Set C:
- From Set B: 1, 1, 2, 4, 7, 7, 8, 8, 10, 10
- Pairing with cards from Set C to sum to 12:
[tex]\[ (4 + 8), (7 + 5 = 12), (8 + 4), (10 + 2), but valid pairs are (4,8) and (10, 2) \][/tex]
Valid winning pairs:
[tex]\[ (4, 8), (10, 2) \][/tex]

We found 2 pairs that sum to 12.

Next, we calculate the total number of possible pairs between Set B and Set C. Since there are 10 cards in Set B and 9 cards in Set C, the total number of pairs is:
[tex]\[ 10 \times 9 = 90 \][/tex]

Now, calculate the probability of winning with Option 2:
[tex]\[ \frac{\text{Number of winning pairs}}{\text{Total number of pairs}} = \frac{2}{90} = 0.0222 \][/tex]

### Conclusion

Comparing the probabilities:
- Probability of winning with Option 1: [tex]\(0.1111 \approx 11.11\%\)[/tex]
- Probability of winning with Option 2: [tex]\(0.0222 \approx 2.22\%\)[/tex]

Option 1 gives a better chance of winning because the probability of winning is higher. Therefore:

[tex]\[\boxed{\text{Option 1}}\][/tex]

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