Questions
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.
(a) Find the value of p and the value of q.
(b) Show that the probability that Francine selects two buttons of the same colour is \(\frac{7}{10}\).
(c) Given that Francine selects two buttons of the same colour, find the probability that she selects two red buttons.
The random variable X is defined as the number of red buttons selected by Francine.
The following table shows the probability distribution of X .
(d) Find the value of a and the value of b .
(e) Hence, find the expected number of red buttons selected by Francine.
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.
(f) Given that the first two buttons she selects are red, write down the probability that the next button she selects is blue.
The probability that she selects the first blue button after n selections in total is \(\frac{3}{56}\).
(g) Find the value of n .
▶️Answer/Explanation
Detailed solution
(a) Find the Value of \( p \) and the Value of \( q \)
The tree diagram represents Francine’s first two selections:
– Initially: 3 red, 1 blue (total 4 buttons).
– First selection:
– Probability of red: \( \frac{3}{4} \),
– Probability of blue: \( \frac{1}{4} \).
– If red is selected, she adds another red, so the bag has 4 red, 1 blue (total 5 buttons).
– Second selection: Probability of red = \( \frac{4}{5} \), blue = \( \frac{1}{5} \).
– If blue is selected, she adds another blue, so the bag has 3 red, 2 blue (total 5 buttons).
– Second selection: Probability of red = \( p \), blue = \( q \).
After selecting blue, the bag has 3 red and 2 blue:
Probability of red: \( p = \frac{3}{5} \),
Probability of blue: \( q = \frac{2}{5} \).
Verify: \( p + q = \frac{3}{5} + \frac{2}{5} = 1 \), which holds.
(b) Show that the Probability Francine Selects Two Buttons of the Same Color is \( \frac{7}{10} \)
Two buttons of the same color means either both red (RR) or both blue (BB):
RR: First red (\( \frac{3}{4} \)), second red (\( \frac{4}{5} \)): \( \frac{3}{4} \cdot \frac{4}{5} = \frac{12}{20} = \frac{3}{5} \).
BB: First blue (\( \frac{1}{4} \)), second blue (\( \frac{2}{5} \)): \( \frac{1}{4} \cdot \frac{2}{5} = \frac{2}{20} = \frac{1}{10} \).
Total probability:
\[
\frac{3}{5} + \frac{1}{10} = \frac{6}{10} + \frac{1}{10} = \frac{7}{10}
\]
This matches the given value.
(c) Given Francine Selects Two Buttons of the Same Color, Find the Probability She Selects Two Red Buttons
Probability of two red buttons given two of the same color:
\[
P(\text{two red} \mid \text{same color}) = \frac{P(\text{two red})}{P(\text{same color})}
\]
\( P(\text{two red}) = \frac{3}{5} \) (from (b)),
\( P(\text{same color}) = \frac{7}{10} \).
\[
\frac{\frac{3}{5}}{\frac{7}{10}} = \frac{3}{5} \cdot \frac{10}{7} = \frac{30}{35} = \frac{6}{7}
\]
(d) Find the Value of \( a \) and the Value of \( b \)
The random variable \( X \) is the number of red buttons selected by Francine (0, 1, or 2):
\( X = 0 \) (BB): \( \frac{1}{4} \cdot \frac{2}{5} = \frac{1}{10} \),
\( X = 1 \) (RB or BR):
RB: \( \frac{3}{4} \cdot \frac{1}{5} = \frac{3}{20} \),
BR: \( \frac{1}{4} \cdot \frac{3}{5} = \frac{3}{20} \),
Total: \( \frac{3}{20} + \frac{3}{20} = \frac{6}{20} = \frac{3}{10} \),
\( X = 2 \) (RR): \( \frac{3}{5} \).
The table gives:
\( P(X = 0) = \frac{1}{10} \), so \( a = 1 \),
\( P(X = 1) = \frac{3}{10} \), so \( b = 3 \).
(e) Hence, Find the Expected Number of Red Buttons Selected by Francine
\[
E(X) = \sum x P(X = x) = 0 \cdot \frac{1}{10} + 1 \cdot \frac{3}{10} + 2 \cdot \frac{6}{10} = 0 + \frac{3}{10} + \frac{12}{10} = \frac{15}{10} = \frac{3}{2}
\]
(f) Given that the First Two Buttons She Selects Are Red, Write Down the Probability that the Next Button She Selects Is Blue
After selecting two reds, the bag has 5 red, 1 blue (total 6 buttons). Probability of blue:
\[
\frac{1}{6}
\]
(g) Find the Value of \( n \)
Francine continues until she selects a blue button. Let’s find the probability of selecting the first blue on the \( n \)-th selection (after \( n-1 \) reds):
– After 1st red: 4 red, 1 blue (5 buttons), blue = \( \frac{1}{5} \).
– After 2nd red: 5 red, 1 blue (6 buttons), blue = \( \frac{1}{6} \).
– After \( k \) reds: \( 3+k \) red, 1 blue, total \( 4+k \), blue = \( \frac{1}{4+k} \).
Probability of first blue on the \( n \)-th selection:
– 1st to \( (n-1) \)-th are red, \( n \)-th is blue:
– 1st red: \( \frac{3}{4} \),
– 2nd red: \( \frac{4}{5} \),
– 3rd red: \( \frac{5}{6} \),
– \( k \)-th red: \( \frac{3+k-1}{4+k-1} = \frac{2+k}{3+k} \),
– \( (n-1) \)-th red: \( \frac{2+(n-1)}{3+(n-1)} = \frac{n+1}{n+2} \),
– \( n \)-th blue: \( \frac{1}{n+3} \).
Product:
\[
\frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdots \frac{n+1}{n+2} \cdot \frac{1}{n+3}
\]
The numerators and denominators cancel:
\[
\frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdots \frac{n+1}{n+2} = \frac{3}{n+2}
\]
\[
P = \frac{3}{n+2} \cdot \frac{1}{n+3} = \frac{3}{(n+2)(n+3)}
\]
Given \( \frac{3}{20} \):
\[
\frac{3}{(n+2)(n+3)} = \frac{3}{20}
\]
\[
(n+2)(n+3) = 20
\]
\[
n^2 + 5n + 6 = 20
\]
\[
n^2 + 5n – 14 = 0
\]
\[
n = \frac{-5 \pm \sqrt{25 + 56}}{2} = \frac{-5 \pm 9}{2}
\]
\[
n = 2 \quad \text{or} \quad n = -7
\]
Since \( n \) is a positive integer, \( n = 2 \).
……………………Markscheme……………………….
Ans:
(a) evidence of understanding that there are now 3R and 2B
\(p=\frac{3}{5}, q=\frac{2}{5}\)
(b) attempt to add two products
P(same) = P(RR or BB) = \(\frac{3}{4}\times \frac{4}{5}+\frac{1}{4}\times \frac{2}{5}\)
=\(\frac{14}{20}\)
=\(\frac{7}{10}\)
(c) attempt to use conditional probability formula in context
\(P(RR|same)=\frac{P(RR)}{P(same)}\)
\(=\frac{\left ( \frac{12}{20} \right )}{\left ( \frac{14}{20} \right )}\)
\(=\frac{12}{14}\left ( =\frac{6}{7} \right )\)
(d) \(a=\frac{6}{20}\left ( =\frac{3}{10} \right ),b=\frac{12}{20}\left ( =\frac{6}{10} \right )\)
(e) attempt to use the formula E(X)
\(E(X)=0\times \frac{1}{10}+1\times \frac{6}{20}+2\times \frac{12}{20}\)
\(=\frac{30}{20}\left ( =\frac{3}{2} \right )\)
(f) \(\frac{1}{6}\)
(g) METHOD 1
METHOD 2
Let X be the number of selections in total made when first blue picked attempt to establish pattern for X=1,2,3,… with at least 3 cases
METHOD 3
\(\frac{3}{4}\times \frac{4}{5}\times \frac{5}{6}\times \frac{6}{7}\times \frac{1}{8}\left ( =\frac{3}{56} \right )\)
so n = 5