Find the term in ${x^5}$ in the expansion of $(3x + A){(2x + B)^6}$.

Mina and Norbert each have a fair cubical die with faces labelled 1, 2, 3, 4, 5 and 6; they throw

it to decide if they are going to eat a cookie.

Mina throws her die just once and she eats a cookie if she throws a four, a five or a six.

Norbert throws his die six times and each time eats a cookie if he throws a five or a six.

Calculate the probability that five cookies are eaten.

$\left( {A\left( \begin{array}{l}6\\5\end{array} \right){2^5}B + 3\left( \begin{array}{l}6\\4\end{array} \right){2^4}{B^2}} \right){x^5}$     M1A1A1

$= \left( {192AB + 720{B^2}} \right){x^5}$     A1

[4 marks]

METHOD 1

$x = \frac{1}{6},{\text{ }}A = \frac{3}{6}\left( { = \frac{1}{2}} \right),{\text{ }}B = \frac{4}{6}\left( { = \frac{2}{3}} \right)$     A1A1A1

probability is $\frac{4}{{81}}{\text{ }}( = 0.0494)$     A1

METHOD 2

P (5 eaten) =P (M eats 1) P (N eats 4) + P (M eats 0) P (N eats 5)     (M1)

$= \frac{1}{2}\left( \begin{array}{l}6\\4\end{array} \right){\left( {\frac{1}{3}} \right)^4}{\left( {\frac{2}{3}} \right)^2} + \frac{1}{2}\left( \begin{array}{l}6\\5\end{array} \right){\left( {\frac{1}{3}} \right)^5}\left( {\frac{2}{3}} \right)$     (A1)(A1)

$= \frac{4}{{81}}{\text{ }}( = 0.0494)$     A1

[4 marks]