A bag contains 3 red and 7 black balls, two balls are taken out at random without replacement. If the first ball taken out is red, then the probability that the second taken out ball is red:

\frac{1}{10}

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\frac{1}{15}

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\frac{3}{10}

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\frac{2}{21}

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Solution

The correct option is C $\frac{1}{15}$
Total no. of balls = 10
P(drawing one red ball) = $\frac{3}{10}$ ( $∵$ red ball = 3)
P(drawing second red ball without replacement) = $\frac{2}{9}$
Required probability $= \frac{3}{10} \times \frac{2}{9} = \frac{1}{15}$

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A bag contains 3 red and 7 black balls, two balls are taken out at random without replacement. If the first ball taken out is red, then the probability that the second taken out ball is red:

The correct option is C 115Total no. of balls = 10 P(drawing one red ball) = 310 (∵ red ball = 3) P(drawing second red ball without replacement) = 29 Required probability =310×29=115

The correct option is C $\frac{1}{15}$
Total no. of balls = 10
P(drawing one red ball) = $\frac{3}{10}$ ( $∵$ red ball = 3)
P(drawing second red ball without replacement) = $\frac{2}{9}$
Required probability $= \frac{3}{10} \times \frac{2}{9} = \frac{1}{15}$