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Question Number 56678 by pieroo last updated on 21/Mar/19
A box contains 9 whites, 7 red and 4 blue  balls. Three balls are picked at random,  one after the other without replacement.  a) find the probability that:  i. they are all of the same color  ii. there is one of each colour  iii. two of them are red  b) if it is known that the first one picked  is blue, find the probability that the rest  are white.
$$\mathrm{A}\:\mathrm{box}\:\mathrm{contains}\:\mathrm{9}\:\mathrm{whites},\:\mathrm{7}\:\mathrm{red}\:\mathrm{and}\:\mathrm{4}\:\mathrm{blue} \\ $$$$\mathrm{balls}.\:\mathrm{Three}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{picked}\:\mathrm{at}\:\mathrm{random}, \\ $$$$\mathrm{one}\:\mathrm{after}\:\mathrm{the}\:\mathrm{other}\:\mathrm{without}\:\mathrm{replacement}. \\ $$$$\left.\mathrm{a}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}: \\ $$$$\mathrm{i}.\:\mathrm{they}\:\mathrm{are}\:\mathrm{all}\:\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{color} \\ $$$$\mathrm{ii}.\:\mathrm{there}\:\mathrm{is}\:\mathrm{one}\:\mathrm{of}\:\mathrm{each}\:\mathrm{colour} \\ $$$$\mathrm{iii}.\:\mathrm{two}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are}\:\mathrm{red} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{the}\:\mathrm{first}\:\mathrm{one}\:\mathrm{picked} \\ $$$$\mathrm{is}\:\mathrm{blue},\:\mathrm{find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{rest} \\ $$$$\mathrm{are}\:\mathrm{white}. \\ $$
Answered by MJS last updated on 21/Mar/19
a.i.  p(www)=((9×8×7)/(20×19×18))=(7/(95))  p(rrr)=((7×6×5)/(20×19×18))=(7/(228))  p(bbb)=((4×3×2)/(20×19×18))=(1/(285))  (7/(95))+(7/(228))+(1/(285))=((41)/(380))≈10.79%  a.ii.  p(wrb)=p(wbr)=p(rwb)=... 6 equal possibilities  6×((9×7×4)/(20×19×18))=((21)/(95))≈22.11%  a.iii.  p(rrx)=p(rxr)=p(xrr) 3 equal possibilities  3×((7×6×13)/(20×19×18))=((91)/(380))≈23.95%  b.  p(ww)=((9×8)/(19×18))=(4/(19))≈21.05%
$$\mathrm{a}.\mathrm{i}. \\ $$$${p}\left(\mathrm{www}\right)=\frac{\mathrm{9}×\mathrm{8}×\mathrm{7}}{\mathrm{20}×\mathrm{19}×\mathrm{18}}=\frac{\mathrm{7}}{\mathrm{95}} \\ $$$${p}\left(\mathrm{rrr}\right)=\frac{\mathrm{7}×\mathrm{6}×\mathrm{5}}{\mathrm{20}×\mathrm{19}×\mathrm{18}}=\frac{\mathrm{7}}{\mathrm{228}} \\ $$$${p}\left(\mathrm{bbb}\right)=\frac{\mathrm{4}×\mathrm{3}×\mathrm{2}}{\mathrm{20}×\mathrm{19}×\mathrm{18}}=\frac{\mathrm{1}}{\mathrm{285}} \\ $$$$\frac{\mathrm{7}}{\mathrm{95}}+\frac{\mathrm{7}}{\mathrm{228}}+\frac{\mathrm{1}}{\mathrm{285}}=\frac{\mathrm{41}}{\mathrm{380}}\approx\mathrm{10}.\mathrm{79\%} \\ $$$$\mathrm{a}.\mathrm{ii}. \\ $$$${p}\left(\mathrm{wrb}\right)={p}\left(\mathrm{wbr}\right)={p}\left(\mathrm{rwb}\right)=…\:\mathrm{6}\:\mathrm{equal}\:\mathrm{possibilities} \\ $$$$\mathrm{6}×\frac{\mathrm{9}×\mathrm{7}×\mathrm{4}}{\mathrm{20}×\mathrm{19}×\mathrm{18}}=\frac{\mathrm{21}}{\mathrm{95}}\approx\mathrm{22}.\mathrm{11\%} \\ $$$$\mathrm{a}.\mathrm{iii}. \\ $$$${p}\left(\mathrm{rrx}\right)={p}\left(\mathrm{rxr}\right)={p}\left(\mathrm{xrr}\right)\:\mathrm{3}\:\mathrm{equal}\:\mathrm{possibilities} \\ $$$$\mathrm{3}×\frac{\mathrm{7}×\mathrm{6}×\mathrm{13}}{\mathrm{20}×\mathrm{19}×\mathrm{18}}=\frac{\mathrm{91}}{\mathrm{380}}\approx\mathrm{23}.\mathrm{95\%} \\ $$$$\mathrm{b}. \\ $$$${p}\left(\mathrm{ww}\right)=\frac{\mathrm{9}×\mathrm{8}}{\mathrm{19}×\mathrm{18}}=\frac{\mathrm{4}}{\mathrm{19}}\approx\mathrm{21}.\mathrm{05\%} \\ $$
Commented by pieroo last updated on 21/Mar/19
Thank you very very much, Sir.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{very}\:\mathrm{much},\:\mathrm{Sir}. \\ $$
Commented by malwaan last updated on 22/Mar/19
a.ii. 6=3!
$$\mathrm{a}.\mathrm{ii}.\:\mathrm{6}=\mathrm{3}! \\ $$

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