A company produces small boxes of sweets that contain 5 jellies and 3 chocolates. Jemeel chooses 3 sweets at random from a box. Draw up the probability distribution table for the number of jellies that Jemeel chooses. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_s20_qp_51 Year：2020 Question No：3(a)
Answer:
$
\begin{array}{l}
\mathrm{P}(0)=\frac{3}{8} \times \frac{2}{7} \times \frac{1}{6}=\frac{1}{56} \\
\mathrm{P}(1)=\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times 3=\frac{15}{56} \\
\mathrm{P}(2)=\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} \times 3=\frac{30}{56} \\
\mathrm{P}(3)=\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6}=\frac{10}{56}
\end{array}
$
(M1 for denominator $8 \times 7 \times 6$ )
Any one probability correct (with correct outcome)
All probabilities correct
\begin{array}{l}
\mathrm{P}(0)=\frac{3}{8} \times \frac{2}{7} \times \frac{1}{6}=\frac{1}{56} \\
\mathrm{P}(1)=\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times 3=\frac{15}{56} \\
\mathrm{P}(2)=\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} \times 3=\frac{30}{56} \\
\mathrm{P}(3)=\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6}=\frac{10}{56}
\end{array}
$
(M1 for denominator $8 \times 7 \times 6$ )
Any one probability correct (with correct outcome)
All probabilities correct
Knowledge points:
5.3.1 evaluate probabilities in simple cases by means of enumeration of equiprobable elementary events, or by calculation using permutations or combinations (e.g. the total score when two fair dice are thrown.) (e.g. drawing balls at random from a bag containing balls of different colours.)
5.4.1 draw up a probability distribution table relating to a given situation involving a discrete random variable X, and calculate E(X) and Var(X)
Solution:
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