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A box contains 25 bulbs out of which 5 are defective. 3 bulbs are to be delivered to a customer. What is the probability that he get one defective bulb?
- a)19/46
- b)25/51
- c)44/77
- d)21/46
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
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A box contains 25 bulbs out of which 5 are defective. 3 bulbs are to b...
5C1*20 C2/25C3
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A box contains 25 bulbs out of which 5 are defective. 3 bulbs are to b...
To find the probability of getting one defective bulb out of three bulbs, we can use the concept of combinations and permutations.
Total number of ways to select 3 bulbs out of 25 bulbs:
This can be calculated using the formula for combinations: nCr = n! / (r!(n-r)!)
Here, n = 25 (total number of bulbs) and r = 3 (number of bulbs to be selected).
So, the total number of ways to select 3 bulbs out of 25 bulbs is given by:
25C3 = 25! / (3!(25-3)!) = 25! / (3!22!) = (25 * 24 * 23) / (3 * 2 * 1) = 2300
Number of ways to select 1 defective bulb and 2 non-defective bulbs:
Since there are 5 defective bulbs and 20 non-defective bulbs, the number of ways to select 1 defective bulb out of 5 and 2 non-defective bulbs out of 20 is given by:
(5C1) * (20C2) = (5! / (1!(5-1)!)) * (20! / (2!(20-2)!)) = (5 * 20 * 19) / (1 * 2) = 190
Probability of getting one defective bulb:
The probability of getting one defective bulb can be calculated by dividing the number of ways to select 1 defective bulb and 2 non-defective bulbs by the total number of ways to select 3 bulbs:
Probability = (Number of ways to select 1 defective bulb and 2 non-defective bulbs) / (Total number of ways to select 3 bulbs)
Probability = 190 / 2300 = 19/46
Therefore, the probability that the customer will get one defective bulb out of three bulbs is 19/46.
Total number of ways to select 3 bulbs out of 25 bulbs:
This can be calculated using the formula for combinations: nCr = n! / (r!(n-r)!)
Here, n = 25 (total number of bulbs) and r = 3 (number of bulbs to be selected).
So, the total number of ways to select 3 bulbs out of 25 bulbs is given by:
25C3 = 25! / (3!(25-3)!) = 25! / (3!22!) = (25 * 24 * 23) / (3 * 2 * 1) = 2300
Number of ways to select 1 defective bulb and 2 non-defective bulbs:
Since there are 5 defective bulbs and 20 non-defective bulbs, the number of ways to select 1 defective bulb out of 5 and 2 non-defective bulbs out of 20 is given by:
(5C1) * (20C2) = (5! / (1!(5-1)!)) * (20! / (2!(20-2)!)) = (5 * 20 * 19) / (1 * 2) = 190
Probability of getting one defective bulb:
The probability of getting one defective bulb can be calculated by dividing the number of ways to select 1 defective bulb and 2 non-defective bulbs by the total number of ways to select 3 bulbs:
Probability = (Number of ways to select 1 defective bulb and 2 non-defective bulbs) / (Total number of ways to select 3 bulbs)
Probability = 190 / 2300 = 19/46
Therefore, the probability that the customer will get one defective bulb out of three bulbs is 19/46.
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