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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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