a) Answer: 0.2 <span>
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) Answer: 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) Answer: 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. </span>
<span>So P(3) + P(4) = 0.5</span>
