3. On his last Friday at work, an IT employee replaces an old PoE switch serving
ID: 2903162 • Letter: 3
Question
3. On his last Friday at work, an IT employee replaces an old PoE switch serving 16 lines with a new one (that also serves 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 came from where, he randomly plugs the 16 lines back into the 16 slots. What is the probability that all of the lines are in the same slots as they were before? What is the probability that none of them are in the same slots as before?
3. On his last Friday at work, an IT employee replaces an old PoE switch serving 16 lines with a new one (that also serves 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 came from where, he randomly plugs the 16 lines back into the 16 slots. What is the probability that all of the lines are in the same slots as they were before? What is the probability that none of them are in the same slots as before?Explanation / Answer
p(none are in same slot) = 1 - p(all are in correct slot)
p(none)= 1 - 1/16!
where 16! denotes 16 factorial( value 16*15*14*13*......2*1)
the value is close to 1 which means that probality of not fitting all of them in same slot as before is very high so the value is approximately 1.
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