A bag contains $3$ blue discs, $2$ red discs and $y$ yellow discs. Li chooses $3$ discs at random fr

Question

A bag contains $3$ blue discs, $2$ red discs and $y$ yellow discs. Li chooses $3$ discs at random from the bag, without replacement. Show that the probability Li chooses $1$ blue disc, $1$ red disc and $1$ yellow disc is $\frac{36y}{\left( y+5 \right)\left( y+4 \right)\left( y+3 \right)}$. [1] Li’s discs are replaced in the bag. Darvina chooses $3$ discs at random from the bag, without replacement. The random variable $S$ is the sum of the number of blue discs chosen and twice the number of red discs chosen. Find an expression in terms of $y$ for $\mathrm{P}\left( S=3 \right)$. [2] Given that $\mathrm{P}\left( S=3 \right)=\frac{7}{20}$, calculate the value of $y$. Hence find the probability distribution of $S$. [6]

Accepted Answer

(a) $\frac{36y}{(y+5)(y+4)(y+3)}$; (b) $\frac{6(6y+1)}{(y+5)(y+4)(y+3)}$; (c) $y = 1$

2025 Specimen Paper P2 Q8

2025 Specimen Paper P2 Q8

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