Question

suppose that 11% of a certain batch of calculators have a defective case, and that 18% have defective batteries. also, 3% have both a defective case and defective batteries. a calculator is selected from the batch at random. find the probability that the calculator has a good case and good batteries. the probability that the calculator has a good case and good batteries is (type an integer or a decimal.)

Explanation:

Step1: Use the principle of inclusion - exclusion

Let $A$ be the event of a defective case and $B$ be the event of defective batteries. We know $P(A)=0.11$, $P(B)=0.18$, and $P(A\cap B) = 0.03$. The probability of a defective case or defective batteries is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
$P(A\cup B)=0.11 + 0.18- 0.03$

Step2: Calculate $P(A\cup B)$

$P(A\cup B)=0.26$

Step3: Find the probability of a good case and good batteries

The probability of a good case and good batteries is the complement of $P(A\cup B)$. Let $P(\text{good})$ be the probability of a good - case and good - batteries. Then $P(\text{good})=1 - P(A\cup B)$.
$P(\text{good})=1 - 0.26$

Answer:

$0.74$