(i)     Find \({\rm{P}}(X = 0)\) .

(ii)     Find \({\rm{P}}(X = 1)\) .

(iii)    Hence, find \({\rm{E}}(X)\) .

Hence, find \({\rm{E}}(X)\) .

Find the probability that exactly five of the balls are gold.

Find the probability that at most five of the balls are gold.

Given that at most five of the balls are gold, find the probability that exactly five of the balls are gold. Give the answer correct to two decimal places.

METHOD 1

(i)     appropriate approach     (M1)

eg   \(\frac{6}{{10}} \times \frac{6}{{10}}\) , \(\frac{6}{{10}} \times \frac{5}{9}\) , \(\frac{6}{{10}} \times \frac{5}{{10}}\)

\({\rm{P}}(X = 0) = \frac{9}{{25}} = 0.36\)     A1     N2

(ii)     multiplying one pair of gold and silver probabilities     (M1)

eg   \(\frac{6}{{10}} \times \frac{4}{{10}}\) , \(\frac{6}{{10}} \times \frac{4}{9}\) , 0.24

adding the product of both pairs of gold and silver probabilities     (M1)

eg   \(\frac{6}{{10}} \times \frac{4}{{10}} \times 2\) , \(\frac{6}{{10}} \times \frac{4}{9} + \frac{4}{{10}} \times \frac{6}{9}\)

\({\rm{P}}(X = 1) = \frac{{12}}{{25}} = 0.48\)     A1     N3

(iii)

\({\rm{P}}(X = 2) = 0.16\) (seen anywhere)     (A1)

correct substitution into formula for \({\rm{E}}(X)\)     (A1)

eg   \(0 \times 0.36 + 1 \times 0.48 + 2 \times 0.16\) , \(0.48 + 0.32\)

\({\rm{E}}(X) = \frac{4}{5} = 0.8\)     A1     N3

METHOD 2

(i)     evidence of recognizing binomial (may be seen in part (ii))     (M1)

eg   \(X \sim {\rm{B}}(2,0.6)\) , \(\left( \begin{array}{l}
2\\
0
\end{array} \right){(0.4)^2}{(0.6)^0}\)

correct probability for use in binomial     (A1)

eg   \(p = 0.4\) , \(X \sim {\rm{B}}(2,0.4)\) , \(^2{C_0}{(0.4)^0}{(0.6)^2}\)

\({\rm{P}}(X = 0) = \frac{9}{{25}} = 0.36\)     A1 N3

(ii)     correct set up     (A1)

eg     \(_2{C_1}{(0.4)^1}{(0.6)^1}\)

\({\rm{P}}(X = 1) = \frac{{12}}{{25}} = 0.48\)     A1     N2

(iii)

attempt to substitute into \(np\)     (M1)

eg   \(2 \times 0.6\)

correct substitution into \(np\)     (A1)

eg   \(2 \times 0.4\)

\({\rm{E}}(X) = \frac{4}{5} = 0.8\)     A1 N3

[8 marks]

METHOD 1

\({\rm{P}}(X = 2) = 0.16\) (seen anywhere)     (A1)

correct substitution into formula for \({\rm{E}}(X)\)     (A1)

eg   \(0 \times 0.36 + 1 \times 0.48 + 2 \times 0.16\) , \(0.48 + 0.32\)

\({\rm{E}}(X) = \frac{4}{5} = 0.8\)     A1     N3

METHOD 2

attempt to substitute into \(np\)     (M1)

eg   \(2 \times 0.6\)

correct substitution into \(np\)     (A1)

eg   \(2 \times 0.4\)

\({\rm{E}}(X) = \frac{4}{5} = 0.8\)     A1 N3

[3 marks]

Let \(Y\) be the number of gold balls drawn from the bag.

evidence of recognizing binomial (seen anywhere)     (M1)

eg   \(_{14}{C_5}{(0.4)^5}{(0.6)^9}\) , \({\rm{B}}(14,0.4)\)

\({\rm{P}}(Y = 5) = 0.207\)     A1     N2

[2 marks]

recognize need to find \({\rm{P}}(Y \le 5)\)     (M1)

\({\rm{P}}(Y \le 5) = 0.486\)     A1     N2

[2 marks]

Let \(Y\) be the number of gold balls drawn from the bag.

recognizing conditional probability     (M1)

eg   \({\rm{P}}(A|B)\) , \({\rm{P}}(Y = 5|Y \le 5)\) , \(\frac{{{\rm{P}}(Y = 5)}}{{{\rm{P}}(Y \le 5)}}\) , \(\frac{{0.207}}{{0.486}}\)

\({\rm{P}}(Y = 5|Y \le 5) = 0.42522518\)     (A1)

\({\rm{P}}(Y = 5|Y \le 5) = 0.43\) (to \(2\) dp)     A1     N2

[3 marks]

Parts (a)(i) and (ii) were generally well done, with candidates either using a tree diagram or a binomial approach. Part (a)(iii) proved difficult, with many either having trouble finding \({\rm{P}}(X = 2)\) or using \({\rm{E}}(X) = np\) .

Part (a)(iii) proved difficult, with many either having trouble finding \({\rm{P}}(X = 2)\) or using \({\rm{E}}(X) = np\) .

A great majority were confident solving part (b) with the GDC, although some did write the binomial term.

Those candidates who did not use the binomial function on the GDC had more difficulty in part (c), although a pleasing number were still able to identify that they were seeking \({\rm{P}}(X \leqslant 5)\) .

While most candidate knew to use conditional probability in part (d), fewer were able to do so successfully, and even fewer still correctly rounded their answer to two decimal places. The most common error was to multiply probabilities to find the intersection needed for the conditional probability formula. Overall, candidates seemed better prepared for probability.