IB Maths AI HL 4.7 Discrete random variables Question Bank

attempt to substitute into the formula for E(I) and recognise that n=2 (or two terms are needed)
$\mathrm{E}(I)=-\frac{1}{10} \ln \left(\frac{1}{10}\right)-\frac{9}{10} \ln \left(\frac{9}{10}\right)$
0.325 (0.325082...)

$\quad \mathrm{E}(I)=-\frac{1}{2} \ln \left(\frac{1}{2}\right)-\frac{1}{2} \ln \left(\frac{1}{2}\right)$
0.693 (0.693147..., $\ln (2)$ )
0.693>0.325

$(\mathrm{E}(I)=)-p \ln p-(1-p) \ln (1-p)$

$\quad(x=) \frac{6}{12} \quad\left(=\frac{1}{2}\right)$

EITHER
for the scales to balance, the odd ball must be in the six balls not chosen

OR
for the scales to balance, all the balls chosen must be of equal weight and hence
$\frac{11}{12} \times \frac{10}{11} \times \frac{9}{10} \times \ldots \times \frac{6}{7}$

EITHER
recognition that the sum of the probabilities on the third row of the table equals 1 e.g. x+2 y=1

OR
for one side to be heavier, the odd ball must be one of six balls chosen $\left(\frac{6}{12}\right)$ and half the time this will result in left-side being heavier, therefore $y=\frac{6}{12} \times \frac{1}{2}$$y=\frac{1}{4}$

$\quad z=-\frac{1}{6} \ln \frac{1}{6}-\frac{1}{6} \ln \frac{1}{6}-\frac{2}{3} \ln \frac{2}{3}$$=0.868 \quad(0.867563 \ldots)$

4 balls on each side because that configuration has the largest E(I)

Note: Award R1 for giving a correct reason AND stating "4 balls on each side"