### Find combinational probability of drawing one color of objects from three different colors by grouping two colors into one large group

Question Sample Titled 'Find combinational probability of drawing one color of objects from three different colors by grouping two colors into one large group'

A box contains ${8}$ red balls , ${4}$ green balls and ${8}$ yellow balls. If ${4}$ balls are randomly drawn from the box at the same time,

 (a) find the probability that exactly ${2}$ red balls are drawn; (2 marks) (b) find the probability that at least ${2}$ red balls are drawn. (2 marks)

 (a) ${P}{\left(\text{exactly two red balls are drawn}\right)}$ $=\dfrac{{{{C}_{{2}}^{{8}}}{{C}_{{2}}^{{12}}}}}{{{{C}_{{4}}^{{20}}}}}$ 1M for numerator $=\dfrac{{616}}{{1615}}$ 1A r.t. ${0.381}$ (b) ${P}{\left(\text{at least two red balls are drawn}\right)}$ $={P}{\left(\text{two red balls}\right)}+{P}{\left(\text{three red balls}\right)}+{P}{\left(\text{four red balls}\right)}$ $=\dfrac{{616}}{{1615}}+\dfrac{{{{C}_{{3}}^{{8}}}{{C}_{{1}}^{{12}}}}}{{{{C}_{{4}}^{{20}}}}}+\dfrac{{{{C}_{{4}}^{{8}}}}}{{{{C}_{{4}}^{{20}}}}}$ 1M for ${\left({a}\right)}+{p}_{{1}}+{p}_{{2}}$ $=\dfrac{{518}}{{969}}$ 1A r.t. ${0.535}$

 (a) ${P}{\left(\text{exactly two red balls are drawn}\right)}$ $={6}{\left(\dfrac{{8}}{{20}}\right)}{\left(\dfrac{{7}}{{19}}\right)}{\left(\dfrac{{12}}{{18}}\right)}{\left(\dfrac{{11}}{{17}}\right)}$ 1M for ${6}$ ${p}_{{3}}{p}_{{4}}{p}_{{5}}{p}_{{6}}$ $=\dfrac{{616}}{{1615}}$ 1A r.t. ${0.381}$   (b) ${P}{\left(\text{at least two red balls are drawn}\right)}$ $={P}{\left(\text{two red balls}\right)}+{P}{\left(\text{three red balls}\right)}+{P}{\left(\text{four red balls}\right)}$ $=\dfrac{{616}}{{1615}}+{4}{\left(\dfrac{{8}}{{20}}\right)}{\left(\dfrac{{7}}{{19}}\right)}{\left(\dfrac{{6}}{{18}}\right)}{\left(\dfrac{{12}}{{17}}\right)}+{\left(\dfrac{{8}}{{20}}\right)}{\left(\dfrac{{7}}{{19}}\right)}{\left(\dfrac{{6}}{{18}}\right)}{\left(\dfrac{{5}}{{17}}\right)}$ 1M for (a)$+{p}_{{7}}+{p}_{{8}}$ $=\dfrac{{518}}{{969}}$ 1A r.t. ${0.535}$   or (b) ${P}{\left(\text{at least two red balls are drawn}\right)}$ $={1}-{P}{\left(\text{no red balls}\right)}-{P}{\left(\text{one red balls}\right)}$ $={1}-\dfrac{{{{C}_{{4}}^{{12}}}}}{{{{C}_{{4}}^{{20}}}}}-\dfrac{{{{C}_{{1}}^{{8}}}{{C}_{{3}}^{{12}}}}}{{{{C}_{{4}}^{{20}}}}}$ 1M for ${1}-{p}_{{9}}-{p}_{{10}}$ $=\dfrac{{518}}{{969}}$ 1A r.t. ${0.535}$

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