Question
A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:
Answer: Option D
$$\eqalign{
& {\text{Here}} \cr
& n\left( E \right){ = ^7}{C_1}{ \times ^5}{C_1}{ \times ^5}{C_1} \cr
& {\text{and,}} \cr
& n\left( S \right){ = ^{12}}{C_1}{ \times ^{11}}{C_1}{ \times ^{10}}{C_1} \cr
& P(S) = \frac{{7 \times 6 \times 5}}{{12 \times 11 \times 10}} \cr
& = \frac{7}{{44}} \cr} $$
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$$\eqalign{
& {\text{Here}} \cr
& n\left( E \right){ = ^7}{C_1}{ \times ^5}{C_1}{ \times ^5}{C_1} \cr
& {\text{and,}} \cr
& n\left( S \right){ = ^{12}}{C_1}{ \times ^{11}}{C_1}{ \times ^{10}}{C_1} \cr
& P(S) = \frac{{7 \times 6 \times 5}}{{12 \times 11 \times 10}} \cr
& = \frac{7}{{44}} \cr} $$
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