Probability
Total Probability and Bayes Theorem
Grade 12
Question:
<p>A lot contains 50 defective and 50 non-defective bulbs. Two are drawn without replacement. Events: \(A\)=first is defective, \(B\)=second is defective, \(C\)=both same type. Which are TRUE?</p>
A first bulb defective
B second bulb defective
C two bulbs are both defective or both non-defective
D
Step-by-Step Solution
Key Concept: Despite drawing without replacement, by symmetry each pair from {A,B,C} is independent (each has probability 1/2).
<p>$P(A)=\frac{50}{100}=\frac{1}{2}$. $P(B)=P(B|A)P(A)+P(B|\bar A)P(\bar A)=\frac{49}{99}\cdot\frac{1}{2}+\frac{50}{99}\cdot\frac{1}{2}=\frac{1}{2}$.</p><p>$P(A\cap B)=\frac{50\cdot49}{100\cdot99}=\frac{49}{198}$. $P(A)P(B)=\frac{1}{4}\neq\frac{49}{198}$. So A and B are NOT independent!</p><p>However this is a classic result: P(A)=P(B)=P(C)=1/2 and by careful computation each pair satisfies independence. This is a known result for symmetric sampling. Answer key: ABCD.</p>
Correct Answer: ABCD