Suppose that two defective refrigerators have been included in a shipment of six refrigerators. The buyer begins to test the six refrigerators one at a time. What is the probability that the last defective refrigerator is found on the fourth test?

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Suppose that two defective refrigerators have been included in a shipment of six refrigerators.   The buyer begins to test the six refrigerators one at a time.   What is the probability that the last defective refrigerator is found on the fourth test?

Szeteib · Accepted Answer

First of all, we denote the events:   G-good refrigerator  D-defective refrigerator  If the last defective refrigerator is found on the 4th test, this means the first defective refrigerator was found on the 1st, 2nd or 3rd test.   So, the possibilities are DGGD, GDGD and GGDD. So, the probability is given, as          P=26⋅45⋅34⋅13+46⋅25⋅34⋅13+46⋅35⋅24⋅13=15  Result: 15

Jeffrey Jordon · Accepted Answer

consider the 4 refrigerators that waschosen out of the 6. There are 6*5*4*3 ways to choose these refrigerators (not 6choose4, since refrigerators are distinguishable). Therefore the denominator of our probability is 360.   The numerator is a bit harder. Consider the first 3 refrigerators that was chosen. 2 were good, 1 was bad. There are 3choose1 = 3 places that the bad refrigerator can go. In each of these places, we count the number of ways to pick these 3 out of the 6. Since one is good, there are 4 ways (b/c there are 4 good refrigerators) to select the first. Another is also good, so the number of ways becomes 4⋅3. The last is bad, and there are 2 bad ones, so the number of ways to select these 3 becomes 4⋅3⋅2. Do not be confused by the fact that I ordered the above, this is taken care of by multiplying by the 3choose1 from above. Think of it like 4⋅3⋅2+4⋅2⋅3+2⋅4⋅3. Regardless, there is only one choice for the 4th refrigerator: bad. Thus the number of ways of being able to find the last bad fridge on the 4th test is 3⋅4⋅3⋅2=72.   The probability is then 72360=0.2  I did the same method for the probability of searching the last bad fridge on the nth trial, and they all added to 1, so I&#039;m pretty sure this is right.

Suppose that two defective refrigerators have been included in a shipment of six refrigerators. The buyer begins to test the six refrigerators one at a time. What is the probability that the last defective refrigerator is found on the fourth test?

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