A batch of 150 toy cars contains 10 toy cars that are defective. Two are selecte
ID: 3208512 • Letter: A
Question
A batch of 150 toy cars contains 10 toy cars that are defective. Two are selected at random from the batch without replacement.
a. What is the probability that the first car selected is non-defective?
b. What is the probability that the first car selected is defective?
c. What is the probability that the second car is defective given that the first car was found defective?
d. What is the probability that the second car is defective given that the first car was found non-defective?
e. If the first car is defective and reintroduced in the batch (still as a defective car), would the probability that the second car is found defective change compared to part c? Demonstrate mathematically.
Explanation / Answer
Total no. cars = 150
No. Defective cars =10; No. Non defective cars = 140
a. Probability of the first car that is being selected is non-defective = No. of non-defective cars /Total no. of cars = 140/150 = 14/15
b. Probability of the first car that is being seleccted defective = No. of defective cars /Total no. of cars = 10/150 = 1/15
c. The probability that the second car is defective given that the first car was found defective
Given that the first car was found defective i.e. The total no. of cars left for the second time selection = 150-1 = 149
The total no. of defective cars left for the second time selection(as he first car is defective) =10-1= 9
Probability that the second car is defective given that the first car was found defective = 9/149
d. the probability that the second car is defective given that the first car was found non-defective
Given that the first car was found defective i.e. The total no. of cars left for the second time selection = 150-1 = 149
But the total no. defective left for the second time selection (=10) are not changed as the first selected car was a non-defective
Probability that the second car is defective given that the first car was found non-defective = 10/149
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