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In a shipment of 20 cars, 3 are found to be defective. If four car s are selected at random, what is the probability that exactly one of the four will be defective?
- a)170/1615
- b)3/20
- c)8/19
- d)3/5
- e)4/5
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
In a shipment of 20 cars, 3 are found to be defective. If four car s a...
There are four possible ways to pick exactly one defective car when picking four cars: DFFF, FDFF, FFDF, FFFD (D = defective, F = functional). To find the total probability we must find the probability of each one of these scenarios and add them together (we add because the total probability is the first scenario OR the second OR…). The probability of the first scenario is the probability of picking a defective car first (3/20) AND then a functional car (17/19) AND then another functional car (16/18) AND then another functional car (15/17). The probability of this first scenario is the product of these four probabilities: 3/20 x 17/19 x 16/18 x 15/17 = 2/19 The probability of each of the other three scenarios would also be 2/19 since the chance of getting the D first is the same as getting it second, third or fourth.
The total probability of getting exactly one defective car out of four = 2/19 + 2/19 + 2/19 + 2/19 = 8/19.
The total probability of getting exactly one defective car out of four = 2/19 + 2/19 + 2/19 + 2/19 = 8/19.
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In a shipment of 20 cars, 3 are found to be defective. If four car s a...
Calculation:
Total number of ways to select 4 cars out of 20:
The total number of ways to select 4 cars out of 20 is given by the combination formula: C(20, 4) = 20! / (4! * (20-4)!) = 4845.
Number of ways to select 1 defective car out of 3:
The number of ways to select 1 defective car out of 3 is simply 3.
Number of ways to select 3 non-defective cars out of 17:
The number of ways to select 3 non-defective cars out of 17 is given by the combination formula: C(17, 3) = 17! / (3! * (17-3)!) = 680.
Probability of selecting exactly 1 defective car:
The probability of selecting exactly 1 defective car out of 4 is given by the number of favorable outcomes divided by the total number of outcomes: (3 * 680) / 4845 = 2040 / 4845 = 8 / 19.
Therefore, the probability that exactly one of the four selected cars will be defective is 8/19, which corresponds to option 'C'.
Total number of ways to select 4 cars out of 20:
The total number of ways to select 4 cars out of 20 is given by the combination formula: C(20, 4) = 20! / (4! * (20-4)!) = 4845.
Number of ways to select 1 defective car out of 3:
The number of ways to select 1 defective car out of 3 is simply 3.
Number of ways to select 3 non-defective cars out of 17:
The number of ways to select 3 non-defective cars out of 17 is given by the combination formula: C(17, 3) = 17! / (3! * (17-3)!) = 680.
Probability of selecting exactly 1 defective car:
The probability of selecting exactly 1 defective car out of 4 is given by the number of favorable outcomes divided by the total number of outcomes: (3 * 680) / 4845 = 2040 / 4845 = 8 / 19.
Therefore, the probability that exactly one of the four selected cars will be defective is 8/19, which corresponds to option 'C'.
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