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

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