A bag contains 7 green and 5 black balls. 3 balls are drawn at random one after the other. If the balls are…

Given,

A bag contains $7$ green and $5$ black balls,

Now $3$ balls are drawn at random one after the other, so number of ways in which we can get green ball without replacement will be ${}^{7}C_1\times {}^{6}C_1\times {}^{5}C_1=7\times 6\times 5$

Now total ways of selecting $3$ balls without replacement will be ${}^{12}C_1\times {}^{11}C_1\times {}^{10}C_1=12\times 11\times 10$

Now the probability of all three balls being green is $\frac{7\times 6\times 5}{12\times 11\times 10}=\frac{7}{44}$