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:

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:

A 123/897

B 23/67

C 7/44

D 12/45

Solution

Correct Answer: Option C

Here, n(E) = \(7C_1\times5C_1\times5C_1\)

And, n(S) = \(12C_1\times11C_1\times10C_1\)

P(S) = \(\frac{7\times6\times5}{12\times11\times10}\) = 7/44