Counting Probability “With Replacement” or “Without Replacement”

If in an experiment both events A and B can happen at once.
Then P(A∩B)=P(A)∙P(B|A)
If events A and B are independent
Then P(A∩B)=P(A)∙P(B)

Example 6: In a box there are 16 lamps, 4 of them are damaged. If two of the lamps are taken randomly one by one without replacement, determine the probability of the 2 taken lamps are damaged.
The explanation:
Suppose A is the event of the first lamp is damaged. The probability of event A is
P(A)=4/16=¼
After the first taking, 15 lamps left, 3 of them are damaged. Suppose B is event of the second taking of damaged lamps.
Probability of event A is P(A)=4/16=¼
Probability of event B is

chance of next event

Probability of the event first taken lamp is damaged and the second withdrawn lamp is damaged equals
P(A∩B)=P(A)∙P(B|A)=¼∙⅕=1/20

Exercise Competency Test 5 of Probability Part III
1. There are 20 dry batteries inside a certain bag, which 5 of them are damaged. If two batteries are taken one by one randomly without replacement then the probability that two taken batterries are damaged =⋯
A. 1/19 B. 2/19 C. 3/19 D. 1/16 E. 1/100
Correct: A, The explanation:
The probability of drawing first damaged battery is P1. The probability of drawing second damaged battery is P2. The probability of drawing two damaged batteries (without replacement) is given by

chance without replacement

2. There are 20 marbles, which are 5 red marbles and 15 black ones. If two marbles are drawn one by one randomly with replacement then the probability of getting 2 red marbles =⋯
A. 1/16 B. ½ C. 1/19 D. 6/16 E. 16/19
Correct: A, The explanation:

probability without replacement

3. There are 20 marbles, which are 5 red marbles and 15 white ones. If two marbles are drawn one by one randomly with replacement then the probability of getting a red at the first and a white at the second =⋯
A. 1/16 B. 2/16 C. 3/16 D. 4/16 E. 5/16
Correct: C, The explanation:

without replacement i

4. A city has 1 fire extinguisher car and 1 ambulance. Probability of the fire extinguisher car can be used when needed is 0.98 and the probability for an ambulance is available when needed is 0.92. If there is an accident caused by fire then the probability of the two cars being available to use =⋯
A. 0.9116 C. 0.8116 D. 0.7116
B. 0.9016 D. 0.8016
Correct: B, The explanation:

P=P1×P2=0.98×0.92=0.9016

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