Three-bags-labelled-R-B-and-W-cotains-Red-Blue-and-White-balls-respectively-of-equal-sizes-the-ratio-of-the-balls-in-the-bag-are-R-B-2-3-and-B-W-4-5-All-the-balls-are-removed-into-a-big

Question Number 176056 by otchereabdullai@gmail.com last updated on 11/Sep/22

$$\:\mathrm{Three}\:\mathrm{bags}\:\mathrm{labelled}\:\mathrm{R},\:\mathrm{B}\:\mathrm{and}\:\mathrm{W}\: \\ $$$$\:\mathrm{cotains}\:\mathrm{Red},\:\mathrm{Blue}\:\mathrm{and}\:\mathrm{White}\:\mathrm{balls}\: \\ $$$$\:\mathrm{respectively}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{sizes},\:\mathrm{the}\:\mathrm{ratio}\: \\ $$$$\:\mathrm{of}\:\mathrm{the}\:\mathrm{balls}\:\mathrm{in}\:\mathrm{the}\:\mathrm{bag}\:\mathrm{are}\:\:\mathrm{R}:\mathrm{B}=\mathrm{2}:\mathrm{3}\: \\ $$$$\mathrm{and}\:\mathrm{B}:\mathrm{W}=\:\mathrm{4}:\mathrm{5}.\:\mathrm{All}\:\mathrm{the}\:\mathrm{balls}\:\mathrm{are}\: \\ $$$$\mathrm{removed}\:\mathrm{into}\:\mathrm{a}\:\mathrm{big}\:\mathrm{bag}\:\mathrm{and}\:\mathrm{properly}\: \\ $$$$\mathrm{mixed}\:\mathrm{together}.\:\mathrm{If}\:\mathrm{two}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{picked} \\ $$$$\mathrm{at}\:\mathrm{random}\:\mathrm{one}\:\mathrm{after}\:\mathrm{the}\:\mathrm{other}\:\mathrm{with}\: \\ $$$$\mathrm{replacement},\:\mathrm{find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\: \\ $$$$\mathrm{picking}\: \\ $$$$\left.\mathrm{a}\right)\:\mathrm{white}\:\mathrm{ball}\:\mathrm{and}\:\mathrm{a}\:\mathrm{blue} \\ $$$$\left.\mathrm{b}\right).\:\mathrm{A}\:\mathrm{blue}\:\mathrm{ball}\:\mathrm{first}\:\mathrm{and}\:\mathrm{then}\:\mathrm{red}\:\mathrm{ball} \\ $$$$ \\ $$

Answered by mr W last updated on 11/Sep/22

(R/B)=(2/3) (B/W)=(4/5) R+B+W=1 ((2B)/3)+B+((5B)/4)=1 ⇒B=((12)/(35)) ⇒R=(2/3)×((12)/(35))=(8/(35)) ⇒W=(5/4)×((12)/(35))=((15)/(35)) i.e. when picking a ball from the big bag, the probability that it′s a red ball is (8/(35)), the probability that it′s a blue ball is ((12)/(35)), the probability that it′s a white ball is ((15)/(35)). a) first white and second blue or first blue and second white p=((15)/(35))×((12)/(35))+((12)/(35))×((15)/(35))=((72)/(245))=0.294 b) first blue and second red p=((12)/(35))×(8/(35))=((96)/(1225))=0.078

$$\frac{{R}}{{B}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\frac{{B}}{{W}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${R}+{B}+{W}=\mathrm{1} \\ $$$$\frac{\mathrm{2}{B}}{\mathrm{3}}+{B}+\frac{\mathrm{5}{B}}{\mathrm{4}}=\mathrm{1} \\ $$$$\Rightarrow{B}=\frac{\mathrm{12}}{\mathrm{35}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{12}}{\mathrm{35}}=\frac{\mathrm{8}}{\mathrm{35}} \\ $$$$\Rightarrow{W}=\frac{\mathrm{5}}{\mathrm{4}}×\frac{\mathrm{12}}{\mathrm{35}}=\frac{\mathrm{15}}{\mathrm{35}} \\ $$$${i}.{e}.\:{when}\:{picking}\:{a}\:{ball}\:{from}\:{the}\:{big} \\ $$$${bag},\:{the}\:{probability}\:{that}\:{it}'{s}\:{a}\:{red}\:{ball} \\ $$$${is}\:\frac{\mathrm{8}}{\mathrm{35}},\:{the}\:{probability}\:{that}\:{it}'{s}\:{a}\:{blue}\:{ball} \\ $$$${is}\:\frac{\mathrm{12}}{\mathrm{35}},\:{the}\:{probability}\:{that}\:{it}'{s}\:{a}\:{white}\:{ball} \\ $$$${is}\:\frac{\mathrm{15}}{\mathrm{35}}. \\ $$$$\left.{a}\right) \\ $$$${first}\:{white}\:{and}\:{second}\:{blue}\:{or} \\ $$$${first}\:{blue}\:{and}\:{second}\:{white} \\ $$$${p}=\frac{\mathrm{15}}{\mathrm{35}}×\frac{\mathrm{12}}{\mathrm{35}}+\frac{\mathrm{12}}{\mathrm{35}}×\frac{\mathrm{15}}{\mathrm{35}}=\frac{\mathrm{72}}{\mathrm{245}}=\mathrm{0}.\mathrm{294} \\ $$$$\left.{b}\right) \\ $$$${first}\:{blue}\:{and}\:{second}\:{red} \\ $$$${p}=\frac{\mathrm{12}}{\mathrm{35}}×\frac{\mathrm{8}}{\mathrm{35}}=\frac{\mathrm{96}}{\mathrm{1225}}=\mathrm{0}.\mathrm{078} \\ $$

Commented by otchereabdullai@gmail.com last updated on 11/Sep/22