Date | May 2011 | Marks available | 3 | Reference code | 11M.2.sl.TZ1.5 |
Level | SL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 5 | Adapted from | N/A |
Question
A box holds 240 eggs. The probability that an egg is brown is 0.05.
Find the expected number of brown eggs in the box.
Find the probability that there are 15 brown eggs in the box.
Find the probability that there are at least 10 brown eggs in the box.
Markscheme
correct substitution into formula for \({\rm{E}}(X)\) (A1)
e.g. \(0.05 \times 240\)
\({\rm{E}}(X) = 12\) A1 N2
[2 marks]
evidence of recognizing binomial probability (may be seen in part (a)) (M1)
e.g. \(\left( {\begin{array}{*{20}{c}}
{240}\\
{15}
\end{array}} \right){(0.05)^{15}}{(0.95)^{225}}\) , \(X \sim {\rm{B}}(240,0.05)\)
\({\rm{P}}(X = 15) = 0.0733\) A1 N2
[2 marks]
\({\rm{P}}(X \le 9) = 0.236\) (A1)
evidence of valid approach (M1)
e.g. using complement, summing probabilities
\({\rm{P}}(X \ge 10) = 0.764\) A1 N3
[3 marks]
Examiners report
Part (a) was answered correctly by most candidates.
In parts (b) and (c), many failed to recognize the binomial nature of this experiment and opted for incorrect techniques in simple probability.
In parts (b) and (c), many failed to recognize the binomial nature of this experiment and opted for incorrect techniques in simple probability. Although several candidates appreciated that (c) involved the idea of a complement, some resorted to elaborate probability addition suggesting they were unaware of the capabilities of their GDC. There was also a great deal of evidence to suggest that candidates did not understand the phrase "at least 10" as several candidates found either \(1 - {\rm{P}}(X \le 10)\) , \(1 - {\rm{P}}(X = 10)\) or \({\rm{P}}(X > 10)\) .