Suppose that two defective refrigerators have been included in a shipment of six

Question

Suppose that two defective refrigerators have been included in a shipment of sixrefrigerators. The buyer begins to test the sixrefrigerators one at a time.

a) What is the probabilitythat the last defective refrigerator is found on the fourth test?

b) What is the probabilitythat no more than four refrigerators need to be tested to locate both of the defective refrigerators?

c) When given that exactlyone of the two defective refrigerators has been located in the rst two tests, what is the probabilitythat the remaining defective refrigerator is found in the third or fourth test?

Explanation / Answer

Lets denote D for defectice and F for functional.

so, total possibilities = DDFFFF, DFDFFF, DFFDFF, DFFFDF, DFFFFD, FDDFFF, FDFDFF, FDFFDF, FDFFFD, FFDDFF, FFDFDF, FFDFFD, FFFDDF, FFFDFD, FFFFDD

a) So, probabilitythat the last defective refrigerator is found on the fourth test = 3/15 = 0.2

b) probability that no more than four refrigerators need to be tested to locate both of the defective refrigerators = 6/15 = 0.4

c) Now, total possibilities = DFDFFF, DFFDFF, DFFFDF, DFFFFD, FDDFFF, FDFDFF, FDFFDF, FDFFFD

Of these 8 cases, 4 of them involve the second refrigerator being in the 3rd or 4th positoin.

Thus, the probability = 4/8 = 0.5

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