$\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) = $\large\frac{20}{81} +$ $\;\large\frac{20}{81} = \frac{40}{81}$

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