A man has 9 identical balls in a bag. Out of these, 3 are black, 2 are blue and the remaining are red.
(a) If a ball is drawn at random, what is the probability that it is (i) not blue? (ii) not red?
(b) If 2 balls are drawn at random, one after the other, what is the probability that both of them will be (i) black, if there is no replacement? (ii) blue, if there is a replacement?
Total number of balls = 9, No of blue = 2, No of black = 3
\(\therefore\) No of red = 4.
(a)(i) P(not blue) = \( 1 - P(blue)\)
= \(1 - \frac{2}{9} = \frac{7}{9}\)
(ii) P(not red) = \(1 - P(red)\)
= \(1 - \frac{4}{9} = \frac{5}{9}\)
(b) Picking two balls;
(i) P(black) with no replacement = \(\frac{3}{9} \times \frac{2}{8} = \frac{1}{12}\)
(ii) P(blue) with replacement = \(\frac{2}{9} \times \frac{2}{9} = \frac{4}{81}\)
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