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# There are two bags,one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls.A die is thrown.If it shows up 1 or 3,a ball is taken from the 1st bag,but it shows up any other number,a ball is chosen from the second bag.Find the probability of choosing a black ball.

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• $E_1=1 \;Bag\; is\; chosen$
• $E_2=2 \;Bag\; is\; chosen$
• $A\;choising\;a\;black\;ball$
• $P(A)=P(E_1)P(A/E_1)+P(E_2)P(A/E_2)$
• $bag\; 1\; is\; chosen\; if \;1 \;or\; 3\; appear \;in \;throw\; of\; die$
• $P(E_1)=\Large\frac{2}{6}=\frac{1}{3}$
• $bag\; 2\; is\; chosen\; otherwise$
• $P(E_1)=1-\Large\frac{1}{3}=\frac{2}{3}$
$P(A/E_1$)=$P(getting\;black\;ball\;/1\;bag\;is\;chosen)$
=$\Large\frac{3}{7}$
$P(A/E_2$)=$P(getting\;black\;ball\;/1\;bag\;is\;chosen)$
=$\Large\frac{4}{7}$
$P(A)=P(getting\;black\;ball$)
=$\Large\frac{1}{3}$$\times$$\frac{3}{7}$+$\frac{2}{3}$$\times$$\frac{4}{7}$
=$\Large\frac{11}{21}$
edited Apr 9, 2014 by pady_1