a) Write multiplicative law of probability for independent events A and B. [1] b) Write conditional probability of event P given that event Q has already occured. [1] c) In a factory producing…

a) Multiplicative Law of Probability for Independent Events

For two independent events $A$ and $B$, the multiplicative law of probability states that the probability of both events occurring simultaneously is the product of their individual probabilities:

$$P(A\cap B)=P(A)\times P(B)$$

Explanation:

• If $A$ and $B$ are independent, the occurrence of one does not affect the probability of the other.
• Example: If $P(A)=0.4$ and $P(B)=0.5$, then $P(A\cap B)=0.4\times 0.5=0.2$.

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b) Conditional Probability of Event $P$ Given $Q$

The conditional probability of event $P$ occurring given that event $Q$ has already occurred is defined as:

$$P(P\mid Q)=\frac{P(P\cap Q)}{P(Q)}$$

Explanation:

• $P(P\cap Q)$: Probability of both $P$ and $Q$ occurring.
• $P(Q)$: Probability of $Q$ occurring.
• Example: If $P(P\cap Q)=0.15$ and $P(Q)=0.3$, then $P(P\mid Q)=\frac{0.15}{0.3}=0.5$.

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c) Probability of Defective Radio from the Third Machine

Given:

• Machine 1: Produces 500 radios/hour, 3% defective.
• Machine 2: Produces 1000 radios/hour, 5% defective.
• Machine 3: Produces 1500 radios/hour, 8% defective.
• A randomly selected defective radio is found.

Find: Probability that the defective radio was produced by the third machine.

Step 1: Calculate Total Production and Defective Radios

• Total radios produced per hour: $500+1000+1500=3000$ radios.

• Defective radios from each machine:

  • Machine 1: $500\times 0.03=15$ defective.
  • Machine 2: $1000\times 0.05=50$ defective.
  • Machine 3: $1500\times 0.08=120$ defective.

• Total defective radios: $15+50+120=185$.

Step 2: Apply Bayes' Theorem

We want $P(\text{Machine 3}\mid \text{Defective})$.

Using the formula: $$P(\text{Machine 3}\mid \text{Defective})=\frac{P(\text{Defective}\mid \text{Machine 3})\times P(\text{Machine 3})}{P(\text{Defective})}$$

• $P(\text{Defective}\mid \text{Machine 3})=0.08$
• $P(\text{Machine 3})=\frac{1500}{3000}=0.5$
• $P(\text{Defective})=\frac{185}{3000}\approx 0.0617$

Substitute the values: $$P(\text{Machine 3}\mid \text{Defective})=\frac{0.08\times 0.5}{0.0617}=\frac{0.04}{0.0617}\approx 0.648$$

Final Answer:

The probability that the defective radio was produced by the third machine is 0.648 (or 64.8%).

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Note: The answer is derived using conditional probability and Bayes' Theorem, ensuring all steps are logically connected.