# 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