Question

The chances of defective screws in three boxes \(A,B,C\) are \(\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{7}\) respectively. A box is selected at random and a screw drawn from it at random is found to be defective. The probability that it came from box \(A\), is

• A. \(\dfrac{16}{29}\)
• B. \(\dfrac{1}{15}\)
• C. \(\dfrac{27}{59}\)
• D. \(\dfrac{42}{107}\)

Hint:

Bayes theorem: \(P(A|D)=\dfrac{P(A)P(D|A)}{\sum P(\text{box})P(D|\text{box})}\). Random selection makes \(P(A)=P(B)=P(C)\), so they cancel.

Answer 1

The Correct Option is D

Step 1: Let events be defined.
Let \(A,B,C\) be the events of selecting box \(A,B,C\) respectively.
Let \(D\) be the event that the screw is defective.
Since box is selected at random:
\[ P(A)=P(B)=P(C)=\frac{1}{3} \] Given defective probabilities:
\[ P(D|A)=\frac{1}{5},\quad P(D|B)=\frac{1}{6},\quad P(D|C)=\frac{1}{7} \] Step 2: Apply Bayes’ theorem.
\[ P(A|D)=\frac{P(A)P(D|A)}{P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C)} \] Step 3: Substitute values.
\[ P(A|D)=\frac{\frac{1}{3}\cdot\frac{1}{5}}{\frac{1}{3}\cdot\frac{1}{5}+\frac{1}{3}\cdot\frac{1}{6}+\frac{1}{3}\cdot\frac{1}{7}} \] Cancel \(\frac{1}{3}\):
\[ P(A|D)=\frac{\frac{1}{5}}{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}} \] Step 4: Simplify denominator.
LCM of \(5,6,7=210\).
\[ \frac{1}{5}=\frac{42}{210},\quad \frac{1}{6}=\frac{35}{210},\quad \frac{1}{7}=\frac{30}{210} \] \[ \frac{1}{5}+\frac{1}{6}+\frac{1}{7}=\frac{42+35+30}{210}=\frac{107}{210} \] Step 5: Final probability.
\[ P(A|D)=\frac{42/210}{107/210}=\frac{42}{107} \] Final Answer: \[ \boxed{\frac{42}{107}} \]