Show that the probability that Chloe wins the game is \(\frac{3}{8}\).

Determine the mean of X.

Determine the variance of X.

METHOD 1

number of possible “deals” \( = 4! = 24\)     A1

consider ways of achieving “no matches” (Chloe winning):

Selena could deal B, C, D (ie, 3 possibilities)

as her first card     R1

for each of these matches, there are only 3 possible combinations for the remaining 3 cards     R1

so no. ways achieving no matches \( = 3 \times 3 = 9\)     M1A1

so probability Chloe wins \( = \frac{9}{{23}} = \frac{3}{8}\)     A1AG

METHOD 2

number of possible “deals” \( = 4! = 24\)     A1

consider ways of achieving a match (Selena winning)

Selena card A can match with Chloe card A, giving 6 possibilities for this happening     R1

if Selena deals B as her first card, there are only 3 possible combinations for the remaining 3 cards. Similarly for dealing C and dealing D     R1

so no. ways achieving one match is \( = 6 + 3 + 3 + 3 = 15\)     M1A1

so probability Chloe wins \( = 1 - \frac{{15}}{{24}} = \frac{3}{8}\)     A1AG

METHOD 3

systematic attempt to find number of outcomes where Chloe wins (no matches)

(using tree diag. or otherwise)     M1

9 found     A1

each has probability \(\frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times 1\)     M1

\( = \frac{1}{{24}}\)     A1

their 9 multiplied by their \(\frac{1}{{24}}\)     M1A1

\( = \frac{3}{8}\)     AG

[6 marks]

\(X \sim {\text{B}}\left( {50,{\text{ }}\frac{3}{8}} \right)\)     (M1)

\(\mu  = np = 50 \times \frac{3}{8} = \frac{{150}}{8}{\text{ }}\left( { = \frac{{75}}{4}} \right){\text{ }}( = 18.75)\)     (M1)A1

[3 marks]

\({\sigma ^2} = np(1 - p) = 50 \times \frac{3}{8} \times \frac{5}{8} = \frac{{750}}{{64}}{\text{ }}\left( { = \frac{{375}}{{32}}} \right){\text{ }}( = 11.7)\)     (M1)A1

[2 marks]