A football team, Melchester Rovers are playing a tournament of five matches.

The probabilities that they win, draw or lose a match are $\frac{1}{2}$, $\frac{1}{6}$ and $\frac{1}{3}$ respectively.

These probabilities remain constant; the result of a match is independent of the results of other matches. At the end of the tournament their coach Roy loses his job if they lose three consecutive matches, otherwise he does not lose his job. Find the probability that Roy loses his job.

METHOD 1

to have $3$ consecutive losses there must be exactly $5$, $4$ or $3$ losses

the probability of exactly $5$ losses (must be $3$ consecutive) is ${\left( {\frac{1}{3}} \right)^5}$     A1

the probability of exactly $4$ losses (with $3$ consecutive) is $4{\left( {\frac{1}{3}} \right)^4}\left( {\frac{2}{3}} \right)$     A1A1

 

Note:     First A1 is for the factor $4$ and second A1 for the other $2$ factors.

 

the probability of exactly $3$ losses (with $3$ consecutive) is $3{\left( {\frac{1}{3}} \right)^3}{\left( {\frac{2}{3}} \right)^2}$     A1A1

 

Note:     First A1 is for the factor $3$ and second A1 for the other $2$ factors.

 

(Since the events are mutually exclusive)

the total probability is $\frac{{1 + 8 + 12}}{{{3^5}}} = \frac{{21}}{{243}}\;\;\;\left( { = \frac{7}{{81}}} \right)$     A1

[6 marks]

METHOD 2

Roy loses his job if

A – first $3$ games are all lost (so the last $2$ games can be any result)

B – first $3$ games are not all lost, but middle $3$ games are all lost (so the first game is not a loss and the last game can be any result)

or C – first $3$ games are not all lost, middle $3$ games are not all lost but last $3$ games are all lost, (so the first game can be any result but the second game is not a loss)

for A ${4^{{\text{th}}}}$ & ${5^{{\text{th}}}}$ games can be anything

${\text{P}}(A) = {\left( {\frac{1}{3}} \right)^3} = \frac{1}{{27}}$     A1

for B ${1^{{\text{st}}}}$ game not a loss & ${5^{{\text{th}}}}$ game can be anything     (R1)

${\text{P}}(B) = \frac{2}{3} \times {\left( {\frac{1}{3}} \right)^3} = \frac{2}{{81}}$     A1

for C ${1^{{\text{st}}}}$ game anything, ${2^{{\text{nd}}}}$ game not a loss     (R1)

${\text{P}}(C) = 1 \times \frac{2}{3} \times {\left( {\frac{1}{3}} \right)^3} = \frac{2}{{81}}$     A1

(Since the events are mutually exclusive)

total probability is $\frac{1}{{27}} + \frac{2}{{81}} + \frac{2}{{81}} = \frac{7}{{81}}$     A1

 

Note:     In both methods all the A marks are independent.

 

Note:     If the candidate misunderstands the question and thinks that it is asking for exactly $3$ losses award A1 A1 and A1 for an answer of $\frac{{12}}{{243}}$ as in the last lines of Method 1.

 

[6 marks]

Total [6 marks]

If a script has lots of numbers with the wrong final answer and no explanation of method it is not going to gain many marks. Working has to be explained. The counting strategy needs to be decided on first. Some candidates misunderstood the context and tried to calculate exactly $3$ consecutive losses. Not putting a non-loss as $\frac{2}{3}$ caused unnecessary work.