Write down

(i)     P(B);
(ii)    P(R).

If the blue face lands down, the die is not rolled again. If the red face lands down, the die is rolled once again. This is represented by the following tree diagram, where p, s, t are probabilities.

Find the value of p, of s and of t.

Guiseppi plays a game where he rolls the die. If a blue face lands down, he scores 2 and is finished. If the red face lands down, he scores 1 and rolls one more time. Let X be the total score obtained.

(i)     Show that \({\text{P}}(X = 3) = \frac{3}{{16}}\) .
(ii)    Find \({\text{P}}(X = 2)\) .

(i)     Construct a probability distribution table for X.

(ii)    Calculate the expected value of X.

If the total score is 3, Guiseppi wins \(\$ 10\). If the total score is 2, Guiseppi gets nothing.

Guiseppi plays the game twice. Find the probability that he wins exactly \(\$ 10\).

(i) P(B) \( = \frac{3}{4}\)     A1     N1

(ii) P(R) \( = \frac{1}{4}\)     A1     N1

[2 marks]

\(p = \frac{3}{4}\)     A1     N1

\(s = \frac{1}{4}\), \(t = \frac{3}{4}\)     A1     N1

[2 marks]

(i) \({\text{P}}(X = 3)\)

\( = {\text{P (getting 1 and 2)}} = \frac{1}{4} \times \frac{3}{4}\)     A1

\( = \frac{3}{{16}}\)     AG     N0

(ii) \({\text{P}}(X = 2) = \frac{1}{4} \times \frac{1}{4} + \frac{3}{4}{\text{ }}\left( {{\text{or }}1 - \frac{3}{{16}}} \right)\)     (A1)

\( = \frac{{13}}{{16}}\)      A1     N2

[3 marks]

(i)

     A2      N2

(ii) evidence of using \({\text{E}}(X) = \sum{x{\text{P}}(X = x)} \)     (M1)
\({\text{E}}(X) = 2\left( {\frac{{13}}{{16}}} \right) + 3\left( {\frac{3}{{16}}} \right)\)    (A1)
\( = \frac{{35}}{{16}}{\text{ }}\left( { = 2\frac{3}{{16}}} \right)\)     A1     N2

[5 marks]

win \(\$ 10 \Rightarrow \) scores 3 one time, 2 other time     (M1)

\({\text{P}}(3) \times {\text{P}}(2) = \frac{{13}}{{16}} \times \frac{3}{{16}}\) (seen anywhere)     A1

evidence of recognising there are different ways of winning \(\$ 10\)     (M1)
e.g. \({\text{P}}(3) \times {\text{P}}(2) + {\text{P}}(2) \times {\text{P}}(3)\) , \(2\left( {\frac{{13}}{{16}} \times \frac{3}{{16}}} \right)\) , \(\frac{{36}}{{256}} + \frac{3}{{256}} + \frac{{36}}{{256}} + \frac{3}{{256}}\)
\({\text{P(win }}\$ 10) = \frac{{78}}{{256}}{\text{ }}\left( { = \frac{{39}}{{128}}} \right)\)     A1     N3

[4 marks]

This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult, with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch. Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to 1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of winning.

This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult, with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch. Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to 1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of winning.

This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult, with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch. Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to 1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of winning.

This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult, with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch. Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to 1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of winning.

This was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult, with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch. Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to 1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the most challenging, but some good answers were seen. The most common error was not recognizing that there were two different ways of winning.