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Home  >>  CBSE XII  >>  Math  >>  Probability
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Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both balls are red. (ii) first ball is black and second is red. (iii) one of them is black and other is red.

$\begin{array}{1 1}(i) 10/18 (ii) 16/81 (iii) 40/80 \\(i) 16/81 (ii) 20/81 (iii) 40/81 \\(i) 4/81 (ii) 10/81 (iii) 20/81 \\ (i) 18/81 (ii) 18/81 (iii) 80/81 \end{array} $

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  • In conditional probability, If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)
Given the total number of black balls = 10 and total number of red balls = 8.
P (drawing a black ball) = $\large \frac{10}{18} = \frac{5}{9}$ and P (drawing a red ball) = $\large\frac{8}{18} = \frac{4}{9}$
(i) P (both balls are red):
P (both balls are red) = P (a red is drawn first and a red is drawn second time also) = P (drawing a red) $\times$ P (drawing a red) = $\large\frac{4}{9}$ $\times$ $\large\frac{4}{9}$ = $\large \frac{16}{81}$
(ii) P (first ball is black and second is red):
P (first ball is black and second is red) = P (drawing a black first) $\times$ P (drawing a red) = $\large\frac{5}{9}$ $\times$ $\large\frac{4}{9}$ = $\large\frac{20}{81}$
(iii) P (getting one black and one red ball):
P (getting one black and one red ball) = P (first ball is black and second is red) + P (first ball is red and second is black)
We calculated the P (drawing a black ball first and red second above) in (ii). The P (drawing a red first and black second) is also the same.
Therefore, P (getting one black and one red ball) = $\large\frac{20}{81} +$ $\;\large\frac{20}{81} = \frac{40}{81}$
answered Jun 19, 2013 by balaji.thirumalai
 

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