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

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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

$$\mathrm{50\%}\:? \\ $$

Commented by Acem last updated on 17/Nov/22

$${It}'{s}\:{less}\:{than}\:{that} \\ $$

Commented by som(math1967) last updated on 17/Nov/22

(((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