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Home  >>  CBSE XI  >>  Math  >>  Probability
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One urn contains two black balls (labelled $B_1,B_2$) and one white ball.A second urn contains one black ball and two white balls (labelled $W_1,W_2$).Suppose the following experiment is performed.One of the two urns is chosen at random.Next a ball is randomly chosen from the urn.Then a second ball is chosen at random from the same urn without replacing the first ball.What is the probability that two balls of opposite colour are chosen.

$\begin{array}{1 1}(A)\;\large\frac{2}{3}\\(B)\;\large\frac{1}{3}\\(C)\;\large\frac{4}{3}\\(D)\;\large\frac{5}{3}\end{array} $

Can you answer this question?
 
 

1 Answer

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Given :
Urn $U_1:B_1,B_2,W$
Urn $U_2:B,W_1,W_2$
Case I:
Selection from first urn
$P=\big[\large\frac{2C_1}{3C_1}\times \large\frac{1}{2C_1}\big]$
$\;\;\;=\big[\large\frac{2}{3}\times \frac{1}{2}\big]$
$\;\;\;=\large\frac{1}{3}$
Step 2:
Case II:
Selection from second urn
$P=\big[\large\frac{1}{3C_1}\times \large\frac{2C_1}{2C_1}\big]$
$\;\;\;=\big[\large\frac{1}{3}\times $$1\big]$
$\;\;\;=\large\frac{1}{3}$
Step 3:
P(Two balls are of opposite colour)=$\large\frac{1}{3}+\frac{1}{3}$
$\Rightarrow \large\frac{2}{3}$
Hence (A) is the correct answer.
answered Jul 8, 2014 by sreemathi.v
 

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