In a stack of cards, 9 cards are blue and the rest are red. If 2 cards are to be...

1. Translate the problem requirements: We have a stack with 9 blue cards and some red cards. When drawing 2 cards without replacement, the probability of both being blue is \(\frac{6}{11}\). We need to find the total number of cards.
2. Set up the probability equation: Express the probability of drawing two blue cards without replacement in terms of the total number of cards.
3. Create and solve the equation: Use the given probability to create an equation and solve for the total number of cards.
4. Verify with answer choices: Check that our solution makes sense and matches one of the given options.

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what we have and what we need to find:

What we know:

• There are 9 blue cards in the stack
• The rest of the cards are red
• We're drawing 2 cards without replacement (once a card is drawn, it's not put back)
• The probability that both cards drawn are blue is \(\frac{6}{11}\)

What we need to find: The total number of cards in the stack

Let's call the total number of cards "n". This means there are (n - 9) red cards.

Process Skill: TRANSLATE - Converting the word problem into mathematical terms we can work with

2. Set up the probability equation

Let's think about this step by step in plain English:

When we draw the first card, what's the chance it's blue? Well, there are 9 blue cards out of n total cards, so the probability is \(\frac{9}{\mathrm{n}}\).

Now, if the first card was blue, what's the chance the second card is also blue? This is where "without replacement" matters. We now have only 8 blue cards left (since we took one), and only (n-1) total cards left. So the probability is \(\frac{8}{\mathrm{n-1}}\).

Since we want BOTH events to happen (first card blue AND second card blue), we multiply these probabilities:

Probability of both blue = \(\frac{9}{\mathrm{n}} \times \frac{8}{\mathrm{n-1}}\)

We know this equals \(\frac{6}{11}\), so:

\(\frac{9}{\mathrm{n}} \times \frac{8}{\mathrm{n-1}} = \frac{6}{11}\)

3. Create and solve the equation

Now let's solve this equation step by step:

\(\frac{9 \times 8}{\mathrm{n} \times (\mathrm{n-1})} = \frac{6}{11}\)

\(\frac{72}{\mathrm{n}(\mathrm{n-1})} = \frac{6}{11}\)

Cross-multiplying: \(72 \times 11 = 6 \times \mathrm{n}(\mathrm{n-1})\)

\(792 = 6\mathrm{n}(\mathrm{n-1})\)

Divide both sides by 6: \(132 = \mathrm{n}(\mathrm{n-1})\)

This gives us: \(\mathrm{n}^2 - \mathrm{n} = 132\)

Rearranging: \(\mathrm{n}^2 - \mathrm{n} - 132 = 0\)

Let's factor this. We need two numbers that multiply to -132 and add to -1. Thinking: 12 and -11 work because \(12 \times (-11) = -132\) and \(12 + (-11) = 1\) So we need: \((\mathrm{n} - 12)(\mathrm{n} + 11) = 0\)

This gives us \(\mathrm{n} = 12\) or \(\mathrm{n} = -11\)

Since we can't have a negative number of cards, n = 12.

Process Skill: MANIPULATE - Systematically solving the quadratic equation and applying constraints

4. Verify with answer choices

Let's check our answer by substituting back:

With 12 total cards and 9 blue cards:

• Probability first card is blue: \(\frac{9}{12} = \frac{3}{4}\)
• Probability second card is blue (given first was blue): \(\frac{8}{11}\)
• Probability both are blue: \(\frac{3}{4} \times \frac{8}{11} = \frac{24}{44} = \frac{6}{11}\) ✓

This matches our given probability of \(\frac{6}{11}\), so our answer is correct.

Looking at the answer choices, 12 corresponds to choice C.

Final Answer

The number of cards in the stack is 12.

Answer: C

Common Faltering Points

Errors while devising the approach

• Misunderstanding "without replacement": Students often confuse this with "with replacement" and incorrectly assume that after drawing the first card, there are still 9 blue cards and n total cards for the second draw. This leads to setting up the probability as \(\frac{9}{\mathrm{n}} \times \frac{9}{\mathrm{n}}\) instead of \(\frac{9}{\mathrm{n}} \times \frac{8}{\mathrm{n-1}}\).
• Incorrect probability setup for sequential events: Some students may think they need to use combinations formula \(\frac{\mathrm{C}(9,2)}{\mathrm{C}(\mathrm{n},2)}\) instead of recognizing this as a sequential probability problem where order matters in the calculation process.
• Misinterpreting the given probability: Students might confuse what \(\frac{6}{11}\) represents - some may think it's the probability of getting exactly one blue card or may set up the equation incorrectly by thinking it represents something other than P(both cards are blue).

Errors while executing the approach

• Cross-multiplication errors: When solving \(\frac{72}{\mathrm{n}(\mathrm{n-1})} = \frac{6}{11}\), students often make arithmetic mistakes in cross-multiplication, getting \(72 \times 6 = 11 \times \mathrm{n}(\mathrm{n-1})\) instead of \(72 \times 11 = 6 \times \mathrm{n}(\mathrm{n-1})\), leading to the wrong quadratic equation.
• Quadratic factoring mistakes: When solving \(\mathrm{n}^2 - \mathrm{n} - 132 = 0\), students may struggle to find the correct factor pair. They need numbers that multiply to -132 and add to -1, but might incorrectly identify factors or make sign errors during factoring.
• Arithmetic errors in simplification: Students may make calculation errors when simplifying \(792 \div 6 = 132\) or when expanding \(\mathrm{n}(\mathrm{n-1})\) to get \(\mathrm{n}^2 - \mathrm{n}\), leading to an incorrect quadratic equation to solve.

Errors while selecting the answer

• Choosing the negative solution: After solving the quadratic equation and getting \(\mathrm{n} = 12\) or \(\mathrm{n} = -11\), some students might not properly apply the constraint that the number of cards must be positive, or they might make an error in determining which solution is valid.
• Skipping verification: Students might arrive at n = 12 but fail to verify their answer by substituting back into the original probability equation, missing the opportunity to catch potential errors in their calculation.