1. How many ways are there to arrange 4 couples if all of the couple must be sta
ID: 2903129 • Letter: 1
Question
1. How many ways are there to arrange 4 couples if all of the couple must be standing together?
2. A collection of 6 items is to be randomly drawn from a bin containing 100 good and 8 defective
items. What is the probability that exactly 3 of the items are defective?
3. On his last Friday at work, an IT employee replaces an old PoE switch serving 16 lines with a new
one (also serving 16 lines.) He mindlessly removes all 16 lines before he realizes that the lines
are not labeled. Instead of taking any time to figure out which line goes where, he randomly
plugs the 16 lines back into the 16 slots. What is the probability that all of the lines are in the
correct slots? What is the probability that none of them are?
4. A drawer contains two boxes of batteries. One box contains 2 good and 8 bad batteries waiting
to be recycled. The other box contains 9 good and 1 bad batteries. Neither box is labeled. An
unsuspecting person opens the drawer and randomly picks one of these boxes, then tests one of
the batteries from the box and it is good. With that knowledge, compute the probability that the
person has the chosen the box with 2 good batteries?
5. A coin is tossed until either 4 heads occur or until the coin has been tossed 7 times. How many
heads/tails sequence are possible? For example, HTHTTHT, HHHH, THHTHH, and TTTTTTT are all
sequence in the list of possible outcomes.
Explanation / Answer
1. Number of ways of arranging 4 couples if the couples must stand together=4!=24.
2. P(defective item is drawn) = 8/108 = 0.074
P(good item is drawn) = 100/108 = 0.925
Let X be the event that a defective item is drawn.
P(X=3) = 6C3 * [p(defective item is drawn)^3] * [p(good item is drawn)^3] =
=>P(X=3) = 20 * 0.0004 * 0.7914 = 0.0063
3. P(All lines are correctly inserted) = 1/16!
P(None of the lines are correctly inserted) = (1-{1/16!})
4. Let P(A) be the probability of choosing the first box with 2 good batteries and P(B) be the probability of choosing the second box with 9 good batteries.
P(A)=1/2=0.5 , P(B)=1/2=0.5
Let P(good)be the probability of choosing a good battery and P(bad) be the probability of choosing a bad battery.
P(good/A) = 2/10 = 1/5 = 0.2 , P(bad/A) = 8/10 = 0.8
P(good/B) = 9/10 = 0.9 , P(bad/B) = 1/10 = 0.1
P(A/good) = [P(good/A) * P(A)] / P(good) = (0.2*0.5)/[(0.5*0.2)+(0.5*0.9)] = 0.18
5. Total number of occurences = [6!/(3!*2!)] + [5!/3!] + [4!/4!] + [(6!*2)/(3!*3!)] + [(6!*2)/(4!*2!)] + [(6!*2)/5!] = 169
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