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. a what is the probability that the last defective refrigerator is found on the fourth test? b what is the probability that no more than four refrigerators need to be tested to locate both of the defective refrigerators? c when given that exactly one of the two defective refrigerators has been located in the first two tests, what is the probability that the remaining defective refrigerator is found in the third or fourth test?

a) 0.2 

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 72/360 = 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'm pretty sure this is right. 

b) 0.4 

This probability is equal to: 
P(find on 2nd) + P(find on 3rd) + P(find on 4th) 

You know the third term, and the first two you can find in the same way as I did above. They are 0.06666... and 0.1333333... 

c) 0.5 

This is P(find on 3rd) + P(find on 4th) 

You know that of the 4 refrigerators left, 3 are good and 1 is bad. So, P(3) = 1/4, because there is a 1 in 4 chance that you choose the bad one right away 

P(4) = 3/4 * 1/3 = 0.5, since there is a 3 in 4 shot you select a good one, then a 1 in 3 that you select a bad one. 

So P(3) + P(4) = 0.5

h/o/e/s mad

step-by-step explanation:

si si si

22.83 repeating

step-by-step explanation:

137 ÷ 6 is 22.8333333 therefore we understand that 3 is repeating.

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