Suppose that 8% of a certain batch of calculators have a defective case, 11% have
defective batteries, and 3% have both of these problems. Construct the probability distribution
table (categories are defective case/non-defective case, and defective batteries/non-defective
batteries). Find the probability that a calculator has
(a) A defective case or defective batteries.
(b) A defective case and good batteries.
(c) Defective batteries, given the case is good.
(d) Is having a defective case independent from having defective batteries?
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85%
a.0.19
b.0.05
c.0.08
d.0.08
To construct the probability distribution table, we need to consider the two categories: defective case (DC) and non-defective case (NDC), and defective batteries (DB) and non-defective batteries (NDB).
First, let's calculate the probabilities for each category:
1) Probability of a defective case (DC): 8%
2) Probability of a non-defective case (NDC): 100% - 8% = 92%
3) Probability of defective batteries (DB): 11%
4) Probability of non-defective batteries (NDB): 100% - 11% = 89%
5) Probability of both defective case and batteries (DC & DB): 3%
Now, let's construct the probability distribution table:
| Defective Case | Non-Defective Case |
---------|----------------|--------------------|
Defective Batteries | 3% | 8% |
Non-Defective Batteries | 8% | 81% |
(a) Probability of a calculator having a defective case or defective batteries:
To find this probability, we need to sum the probabilities in the table that fall under the "defective case" or "defective batteries" categories:
P(defective case or defective batteries) = P(DC) + P(DB) - P(DC & DB)
P(defective case or defective batteries) = 8% + 11% - 3% = 16%
(b) Probability of a calculator having a defective case and good batteries:
To find this probability, we need to find the intersection of the "defective case" and "non-defective batteries" categories:
P(defective case and good batteries) = P(DC) - P(DC & DB)
P(defective case and good batteries) = 8% - 3% = 5%
(c) Probability of defective batteries given the case is good:
To find this probability, we need to consider the conditional probability of "defective batteries" given "non-defective case":
P(defective batteries | good case) = P(DB) / P(NDC)
P(defective batteries | good case) = 11% / 92% = 0.1196 = 11.96%
(d) Independence of defective case and defective batteries:
Two events are independent if knowing the occurrence of one event does not affect the occurrence of the other event. To determine if the two events are independent, we need to compare the joint probability to the product of the individual probabilities.
P(DC & DB) = 3%
P(DC) * P(DB) = 8% * 11% = 0.088% = 8.8%
Since P(DC & DB) ≠ P(DC) * P(DB), the defective case and defective batteries are not independent.