Find the probability of obtaining exactly two heads.

Find the probability of obtaining at least two heads.

evidence of using binomial probability     (M1)

e.g. \({\rm{P}}(X = 2) = \left( {\begin{array}{*{20}{c}}
7\\
2
\end{array}} \right){(0.18)^2}{(0.82)^5}\)

\({\rm{P}}(X = 2) = 0.252\)     A1     N2

[2 marks]

METHOD 1

evidence of using the complement     M1

e.g. \(1 - ({\rm{P}}(X \le 1))\)

\({\text{P}}(X \le 1) = 0.632\)    (A1)

\({\text{P}}(X \ge 2) = 0.368\)    A1     N2

METHOD 2

evidence of attempting to sum probabilities     M1

e.g. \({\text{P(2 heads) + P(3 heads)}} + \ldots + {\text{P(7 heads)}}\) , \(0.252 + 0.0923 + \ldots \)

correct values for each probability     (A1)

e.g. \(0.252 + 0.0923 + 0.0203 + 0.00267 + 0.0002 + 0.0000061\)

\({\text{P(}}X \ge {\rm{2) = 0}}{\rm{.368}}\)     A1     N2

[3 marks]

Candidates who recognized binomial probability answered this question very well, using their GDC to perform the final calculations.

Some candidates misinterpreted the meaning of "at least two" in part (b), and instead found \({\text{P(}}X > {\rm{2)}}\) . Others wrote down a correct interpretation but accumulated to in their GDC (e.g. binomcdf (7, 0.18, 2)). Still, the number of candidates who either left this question blank or approached the question without binomial considerations suggests that this topic continues to be neglected in some centres.