Question

There are 10 machines of which exactly 3 are defective. The machines are tested one by one until all defective are identified. Then the probability that the last defective is tested in 6th attempt is

• A.
  3. $\frac { { } ^ { 7 } \mathrm { C } _ { 3 } } { { } ^ { 10 } \mathrm { C } _ { 5 } }$
• B. $\frac { 3 } { 5 } \cdot \frac { { } ^ { 7 } \mathrm { C } _ { 3 } } { { } ^ { 10 } \mathrm { C } _ { 5 } }$
• C. $\frac { 1 } { 5 } \cdot \frac { { } ^ { 7 } \mathrm { C } _ { 3 } } { { } ^ { 10 } \mathrm { C } _ { 5 } }$
• D.

Answer 1

Answer: (B)

$\frac { 3 } { 5 } \cdot \frac { { } ^ { 7 } \mathrm { C } _ { 3 } } { { } ^ { 10 } \mathrm { C } _ { 5 } }$

Answer 2

N(S) = ¹⁰C₅ n(5) ³C₂ ⁷C₃

P(5) = $\frac { { } ^ { 3 } \mathrm { C } _ { 2 } \cdot { } ^ { 7 } \mathrm { C } _ { 3 } } { { } ^ { 10 } \mathrm { C } _ { 5 } }$ ; P(E₁/E) = 1/5

∴ P(E∩E₁) = P(5). P(E₁/E).