As soon as Sarah misses a total of 4 lessons at her school an email is sent to her parents. The probability that she misses any particular lesson is constant with a value of \(\frac{1}{3}\). Her decision to attend a lesson is independent of her previous decisions.

(a)     Find the probability that an email is sent to Sarah’s parents after the \({8^{{\text{th}}}}\) lesson that Sarah was scheduled to attend.

(b)     If an email is sent to Sarah’s parents after the \({X^{{\text{th}}}}\) lesson that she was scheduled to attend, find \({\text{E}}(X)\).

(c)     If after 6 of Sarah’s scheduled lessons we are told that she has missed exactly 2 lessons, find the probability that an email is sent to her parents after a total of 12 scheduled lessons.

(d)     If we know that an email was sent to Sarah’s parents immediately after her \({6^{{\text{th}}}}\) scheduled lesson, find the probability that Sarah missed her \({2^{{\text{nd}}}}\) scheduled lesson.

(a)     we are dealing with the Negative Binomial distribution: \({\text{NB}}\left( {4,\frac{1}{3}} \right)\)     (M1)

let X be the number of scheduled lessons before the email is sent

\({\text{P}}(X = 8) = \left( {\begin{array}{*{20}{c}}
  7 \\
  3
\end{array}} \right){\left( {\frac{2}{3}} \right)^4}{\left( {\frac{1}{3}} \right)^4} = 0.0854\)     (M1)A1

[3 marks]

(b)     \({\text{E}}(X) = \frac{r}{p} = \frac{4}{{\frac{1}{3}}} = 12\)     (M1)A1

[2 marks]

(c)     we are asking for 2 missed lessons in the second 6 lessons, with the last lesson missed so this is \({\text{NB}}\left( {2,\frac{1}{3}} \right)\)     (M1)

\({\text{P}}(X = 6) = \left( {\begin{array}{*{20}{c}}
  5 \\
  1
\end{array}} \right){\left( {\frac{2}{3}} \right)^4}{\left( {\frac{1}{3}} \right)^2} = 0.110\)     (M1)A1

Note: Accept solutions laid out in terms of conditional probabilities.

[3 marks]

(d)     EITHER

We know that she missed the \({6^{{\text{th}}}}\) lesson so she must have missed 3 from the first 5 lessons. All are equally likely so the probability that she missed the \({2^{{\text{nd}}}}\) lesson is \(\frac{3}{5}\).     R1A1

OR

require \({\text{P(missed }}{{\text{2}}^{{\text{nd}}}}|X = 6) = \frac{{{\text{P(missed }}{{\text{2}}^{{\text{nd}}}}{\text{ and }}X = 6)}}{{{\text{P}}(X = 6)}}\)     R1

\({\text{P(missed }}{{\text{2}}^{{\text{nd}}}}{\text{ and }}X = 6) = {\text{P(missed }}{{\text{2}}^{{\text{nd}}}}{\text{ and }}{{\text{6}}^{{\text{th}}}}{\text{ and 2 of remaining 4)}}\)

\[ = \frac{1}{3} \cdot \frac{1}{3} \cdot \left( {\begin{array}{*{20}{c}}
  4 \\
  2
\end{array}} \right){\left( {\frac{1}{3}} \right)^2}{\left( {\frac{2}{3}} \right)^2} = \frac{{24}}{{{3^6}}}\]

\({\text{P}}(X = 6) = \left( {\begin{array}{*{20}{c}}
  5 \\
  3
\end{array}} \right){\left( {\frac{1}{3}} \right)^4}{\left( {\frac{2}{3}} \right)^2} = \frac{{40}}{{{3^6}}}\)

so required probability is \(\frac{{24}}{{{3^6}}} \cdot \frac{{{3^6}}}{{40}} = \frac{3}{5}\)     A1

[2 marks]

Total [10 marks]

Realising that this was a problem about the Negative Binomial distribution was the crucial thing to realise in this question. All parts of the syllabus do need to be covered.