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IB Mathematics SL 4.1 Concepts of population, sample AA SL Paper 1- Exam Style Questions

IB Mathematics SL 4.1 Concepts of population, sample AA SL Paper 1- Exam Style Questions- New Syllabus

IB DP Maths AA SL Paper 1- All Topics
Question

A bag contains buttons which are either red or blue. Initially, the bag contains three red buttons and one blue button. Francine randomly selects one button from the bag. She then replaces the button and adds one extra button of the same colour. For example, if she selects a red button, she then replaces it and adds one extra red button so that the bag then contains four red buttons and one blue button. Francine then randomly selects a second button from the bag. The following tree diagram represents the probabilities of the first two selections.

Tree diagram for button selections

Part (a):
Find the value of \( p \) and the value of \( q \). [2]

Part (b):
Show that the probability that Francine selects two buttons of the same colour is \( \frac{7}{10} \). [3]

Part (c):
Given that Francine selects two buttons of the same colour, find the probability that she selects two red buttons. [2]

Part (d):
The random variable \( X \) is defined as the number of red buttons selected by Francine. The following table shows the probability distribution of \( X \).
Probability distribution table
Find the value of \( a \) and the value of \( b \). [2]

Part (e):
Hence, find the expected number of red buttons selected by Francine. [2]

Part (f):
Francine restarts the process with three red buttons and one blue button in the bag. She selects buttons as before, replacing the button and adding one extra button of the same colour each time. She repeats this until she selects a blue button. Given that the first two buttons she selects are red, write down the probability that the next button she selects is blue. [1]

Part (g):
The probability that she selects the first blue button after \( n \) selections in total is \( \frac{3}{56} \). Find the value of \( n \). [3]

▶️ Answer/Explanation
Solutions

Part (a)

Find the value of \( p \) and \( q \).

Initial bag: 3 red, 1 blue (4 buttons).

First selection: \( P(\text{red}) = \frac{3}{4} \), \( P(\text{blue}) = \frac{1}{4} \).

If blue selected, add blue: 3 red, 2 blue (5 buttons).

Second selection: \( p = P(\text{red}) = \frac{3}{5} \), \( q = P(\text{blue}) = \frac{2}{5} \).

Verify: \( \frac{3}{5} + \frac{2}{5} = 1 \).

Answer: \( p = \frac{3}{5} \), \( q = \frac{2}{5} \)

Part (b)

Show the probability of two buttons of the same colour is \( \frac{7}{10} \).

RR: \( \frac{3}{4} \cdot \frac{4}{5} = \frac{12}{20} = \frac{3}{5} \)

BB: \( \frac{1}{4} \cdot \frac{2}{5} = \frac{2}{20} = \frac{1}{10} \)

Total: \( \frac{3}{5} + \frac{1}{10} = \frac{6}{10} + \frac{1}{10} = \frac{7}{10} \)

Answer: Probability is \( \frac{7}{10} \)

Part (c)

Probability of two red buttons given two of the same colour:

\[ P(\text{RR} \mid \text{same}) = \frac{P(\text{RR})}{P(\text{same})} = \frac{\frac{3}{5}}{\frac{7}{10}} = \frac{3}{5} \cdot \frac{10}{7} = \frac{6}{7} \]

Answer: \( \frac{6}{7} \)

Part (d)

Find \( a \) and \( b \) in the probability distribution of \( X \).

\( X = 0 \) (BB): \( \frac{1}{4} \cdot \frac{2}{5} = \frac{1}{10} \)

\( X = 1 \) (RB or BR): \( \frac{3}{4} \cdot \frac{1}{5} + \frac{1}{4} \cdot \frac{3}{5} = \frac{3}{20} + \frac{3}{20} = \frac{6}{20} \)

\( X = 2 \) (RR): \( \frac{3}{4} \cdot \frac{4}{5} = \frac{12}{20} \)

Table: \( P(X = 1) = \frac{a}{20} = \frac{6}{20} \), so \( a = 6 \).

\( P(X = 2) = \frac{b}{20} = \frac{12}{20} \), so \( b = 12 \).

Answer: \( a = 6 \), \( b = 12 \)

Part (e)

Expected number of red buttons:

\[ E(X) = 0 \cdot \frac{1}{10} + 1 \cdot \frac{6}{20} + 2 \cdot \frac{12}{20} = \frac{3}{10} + \frac{24}{20} = \frac{3}{10} + \frac{12}{10} = \frac{15}{10} = \frac{3}{2} \]

Answer: \( \frac{3}{2} \)

Part (f)

Probability of blue after two reds:

After 1st red: 4 red, 1 blue (5 buttons).

After 2nd red: 5 red, 1 blue (6 buttons).

Third selection: \( P(\text{blue}) = \frac{1}{6} \).

Answer: \( \frac{1}{6} \)

Part (g)

Find \( n \) where \( P(\text{first blue on } n\text{-th selection}) = \frac{3}{56} \).

After \( n-1 \) reds: \( 3 + (n-1) \) red, 1 blue, total \( 4 + (n-1) = n + 3 \).

Probability: \( \frac{3}{4} \cdot \frac{4}{5} \cdot \ldots \cdot \frac{n+1}{n+2} \cdot \frac{1}{n+3} = \frac{3}{n+2} \cdot \frac{1}{n+3} \).

\[ \frac{3}{(n+2)(n+3)} = \frac{3}{56} \]

\[ (n+2)(n+3) = 56 \]

\[ n^2 + 5n + 6 = 56 \]

\[ n^2 + 5n – 50 = 0 \]

\[ n = \frac{-5 \pm \sqrt{25 + 200}}{2} = \frac{-5 \pm 15}{2} \]

\[ n = 5 \text{ or } n = -10 \]

Since \( n > 0 \), \( n = 5 \).

Answer: \( n = 5 \)

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