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

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Question Number 56678 by pieroo last updated on 21/Mar/19

$$\mathrm{A}\mathrm{box}\mathrm{contains}9\mathrm{whites},7\mathrm{red}\mathrm{and}4\mathrm{blue}$$$$\mathrm{balls}.\mathrm{Three}\mathrm{balls}\mathrm{are}\mathrm{picked}\mathrm{at}\mathrm{random},$$$$\mathrm{one}\mathrm{after}\mathrm{the}\mathrm{other}\mathrm{without}\mathrm{replacement}.$$$$\mathrm{a})\mathrm{find}\mathrm{the}\mathrm{probability}\mathrm{that}:$$$$\mathrm{i}.\mathrm{they}\mathrm{are}\mathrm{all}\mathrm{of}\mathrm{the}\mathrm{same}\mathrm{color}$$$$\mathrm{ii}.\mathrm{there}\mathrm{is}\mathrm{one}\mathrm{of}\mathrm{each}\mathrm{colour}$$$$\mathrm{iii}.\mathrm{two}\mathrm{of}\mathrm{them}\mathrm{are}\mathrm{red}$$$$\mathrm{b})\mathrm{if}\mathrm{it}\mathrm{is}\mathrm{known}\mathrm{that}\mathrm{the}\mathrm{first}\mathrm{one}\mathrm{picked}$$$$\mathrm{is}\mathrm{blue},\mathrm{find}\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{rest}$$$$\mathrm{are}\mathrm{white}.$$

Answered by MJS last updated on 21/Mar/19

$$\mathrm{a}.\mathrm{i}.$$$$p\left(\mathrm{www}\right)=\frac{9\times 8\times 7}{20\times 19\times 18}=\frac{7}{95}$$$$p\left(\mathrm{rrr}\right)=\frac{7\times 6\times 5}{20\times 19\times 18}=\frac{7}{228}$$$$p\left(\mathrm{bbb}\right)=\frac{4\times 3\times 2}{20\times 19\times 18}=\frac{1}{285}$$$$\frac{7}{95}+\frac{7}{228}+\frac{1}{285}=\frac{41}{380}\approx 10.79\mathrm{\%}$$$$\mathrm{a}.\mathrm{ii}.$$$$p\left(\mathrm{wrb}\right)=p\left(\mathrm{wbr}\right)=p\left(\mathrm{rwb}\right)=\dots 6\mathrm{equal}\mathrm{possibilities}$$$$6\times \frac{9\times 7\times 4}{20\times 19\times 18}=\frac{21}{95}\approx 22.11\mathrm{\%}$$$$\mathrm{a}.\mathrm{iii}.$$$$p\left(\mathrm{rrx}\right)=p\left(\mathrm{rxr}\right)=p\left(\mathrm{xrr}\right)3\mathrm{equal}\mathrm{possibilities}$$$$3\times \frac{7\times 6\times 13}{20\times 19\times 18}=\frac{91}{380}\approx 23.95\mathrm{\%}$$$$\mathrm{b}.$$$$p\left(\mathrm{ww}\right)=\frac{9\times 8}{19\times 18}=\frac{4}{19}\approx 21.05\mathrm{\%}$$

Commented by pieroo last updated on 21/Mar/19

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{very}\mathrm{much},\mathrm{Sir}.$$

Commented by malwaan last updated on 22/Mar/19

$$\mathrm{a}.\mathrm{ii}.6=3!$$

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