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The probability of drawing two black balls in succession form a bag containing 4- red,3- black balls when the ball that drawn first in replaced is

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$\frac{1}{49}$

$\frac{^{9} C_{1}}{^{49} C_{1}}$

$\frac{^{40} C_{39}}{^{40} C_{38}}$

$\frac{40}{49}$

answer is C.

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Detailed Solution

$n (S) = 4 + 3 = 7$

Required probability $= \frac{^{3} C_{1} \times^{3} C_{1}}{^{7} C_{1} \times^{7} C_{1}} = \frac{3 \times 3}{7 \times 7}$

$= \frac{9}{49}$

$= \frac{^{9} C_{1}}{^{49} C_{1}}$

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