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There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are           randomly picked for inspection (i.e., each has the same chance of being selected), what is the probability that only one of the defective calculators will be included in the inspection?
  • a)
    1/2
  • b)
    1/3
  • c)
    1/4
  • d)
    1/5
Correct answer is option 'B'. Can you explain this answer?
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There are 25 calculators in a box. Two of them are defective. Suppose ...
Probability of only one is defective out of 5 calculators 
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There are 25 calculators in a box. Two of them are defective. Suppose ...
Problem:
There are 25 calculators in a box, out of which 2 are defective. Suppose 5 calculators are randomly picked for inspection. What is the probability that only one of the defective calculators will be included in the inspection?

Solution:
To find the probability that only one of the defective calculators will be included in the inspection, we need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.

Favorable Outcomes:
To have only one defective calculator included in the inspection, we can choose one defective calculator out of the two available, and then choose four non-defective calculators from the remaining 23. The number of ways to do this can be calculated using the combination formula.

Number of ways to choose 1 defective calculator from 2: C(2, 1) = 2
Number of ways to choose 4 non-defective calculators from 23: C(23, 4) = 23! / (4! * 19!) = 8855

So, the number of favorable outcomes is 2 * 8855 = 17710.

Total Outcomes:
To find the total number of possible outcomes, we need to choose any 5 calculators from the total 25 available. Again, this can be calculated using the combination formula.

Number of ways to choose 5 calculators from 25: C(25, 5) = 25! / (5! * 20!) = 53130

So, the total number of possible outcomes is 53130.

Probability:
Now, we can find the probability by dividing the number of favorable outcomes by the total number of possible outcomes.

Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 17710 / 53130
Probability = 1/3

Therefore, the probability that only one of the defective calculators will be included in the inspection is 1/3. Hence, option B is the correct answer.
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