Find the expected number of questions Jose answers correctly.

Find the probability that Jose answers exactly three questions correctly.

Find the probability that Jose answers more than three questions correctly.

\({\rm{E}}(X) = 2\)     A1     N1

[1 mark]

evidence of appropriate approach involving binomial     (M1)

e.g.  \(\left( {\begin{array}{*{20}{c}}
{10}\\
3
\end{array}} \right){(0.2)^3}\) , \({(0.2)^3}{(0.8)^7}\) , \(X \sim {\rm{B}}(10,0.2)\)

\({\rm{P}}(X = 3) = 0.201\)     A1     N2

[2 marks]

METHOD 1

\({\rm{P}}(X \le 3) = 0.10737 + 0.26844 + 0.30199 + 0.20133\) \(( = 0.87912 \ldots )\)     (A1)

evidence of using the complement (seen anywhere)     (M1)

e.g. \(1 - \) any probability , \({\rm{P}}(X > 3) = 1 - {\rm{P}}(X \le 3)\)

\({\rm{P}}(X > 3) = 0.121\)     A1     N2

METHOD 2

recognizing that \({\rm{P}}(X > 3) = {\rm{P}}(X \ge 4)\)     (M1)

e.g. summing probabilities from \(X = 4\) to \(X = 10\)

correct expression or values     (A1)

e.g. \(\sum\limits_{r = 4}^{10} {\left( {\begin{array}{*{20}{c}}
{10}\\
r
\end{array}} \right)} {(0.2)^{10 - r}}{(0.8)^r}\)

\(0.08808 + 0.02642 + 0.005505 + 0.000786 + 0.0000737 + 0.000004 + 0.0000001\)

\({\rm{P}}(X > 3) = 0.121\)     A1     N2

[3 marks]

Most candidates were able to find the mean by applying various methods. Although many recognised binomial probability, fewer were able to use the GDC effectively.

Most candidates were able to find the mean by applying various methods. Although many recognised binomial probability, fewer were able to use the GDC effectively.

Part (c) was problematic in some cases but most candidates recognized that either a sum of probabilities or the complement was required. Many misinterpreted "more than three" as inclusive of three, and so obtained incorrect answers. When adding individual probabilities, some candidates used three or fewer significant figures, which resulted in an incorrect final answer due to premature rounding.