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Question Number 180775 by Acem last updated on 17/Nov/22
$${A}\:{box}\:{contains}\:\mathrm{2}\:{blue},\:\mathrm{1}\:{red}\:{balls}.\:{We}\:{randomly} \\ $$$$\:{draw}\:{one}\:{ball}\:{then}\:{put}\:{it}\:{back}\:{and}\:{add}\:{two}\:{balls} \\ $$$$\:{of}\:{the}\:{same}\:{color}\:{of}\:{that}\:{ball}\:{in}\:{the}\:{box}.\:{We} \\ $$$$\:{randomly}\:{draw}\:{again}\:{one}\:{ball},\:{if}\:{it}\:{was}\:{red}\:{then} \\ $$$$\:{what}'{s}\:{the}\:{probability}\:{that}\:{the}\:{first}\:{drawn}\:{ball} \\ $$$$\:{was}\:{blue}? \\ $$
Commented by som(math1967) last updated on 17/Nov/22
$$\mathrm{50\%}\:? \\ $$
Commented by Acem last updated on 17/Nov/22
$${It}'{s}\:{less}\:{than}\:{that} \\ $$
Commented by som(math1967) last updated on 17/Nov/22
$$\:\:\:\frac{\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{5}}}{\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{5}}}=\frac{\mathrm{2}}{\mathrm{15}}×\frac{\mathrm{15}}{\mathrm{5}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\:{i}.{e}\:\mathrm{40\%} \\ $$
Commented by Acem last updated on 17/Nov/22
$${Correct}\:{Sir}!\:{Thanks}! \\ $$
Commented by som(math1967) last updated on 17/Nov/22
$${I}\:{made}\:{a}\:{mistake}\:{in}\:{the}\:{calculation} \\ $$
Commented by Acem last updated on 17/Nov/22
$${What}\:{matters}\:{is}\:{the}\:{method},\:{and}\:{yours}\:{was} \\ $$$$\:\mathrm{100\%}\:{correct}!\:{thank}\:{you}\:{very}\:{much}. \\ $$
Commented by SLVR last updated on 18/Nov/22
$${Good}\:{sir} \\ $$