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bahrom7893
Four students place their phones on a desk. Later they each pick up a phone at random. What is the probability that exactly one student gets his/her phone?
@satellite73 can u help me out.
Arnab that's wrong. The probability that the first student gets his phone is 1/4th. What if the 2nd guy gets the phone. Then one phone is gone.
So the probability that second guy gets it is 1/3
ash, the answer's 1/3
idk how they got it though. Exactly one student gets the phone back.
100 % probability if they are in senses. lol but it ll be 1/24
oh sorry, lemme think..
can someone please help me? x :)
yeah, it is coming 1/3 :)
I'm thinking: (1/4) "first student got" + (2nd|Nobody else got theirs)
and then 3rd got his given nobody else got theirs and then 4th got his, etc..
first, pick up a phone at random and give it to the right person. it can be done in 4C1 ways= 4 ways. rest of the 3 are to be deranged, so, D3= 2 ways. so, the condition can be satisfied in total 4*2 ways=8 ways out of 4! ways= 24 ways. so, the probability is 8/24=1/3
what do u mean by deranged? this may be a dumb question, but I've never heard of the term before.
u know about derangement?
ok, derangement is a kind of arrangement where no right thing goes to right place..
Ohh ok. What is the formula for derangement?
there is a general method: Dr= r!(1/2!-1/3!+1/4!-..... upto 1/n!)
This problem is an updated version of the Secretary's Packet Problem A secretary types four letters to four people and addresses the four envelopes. If she inserts the letters at random, each in a different envelope, what is the probability that exactly three letters will go into the right envelope? http://www.cut-the-knot.org/Probability/IntuitiveProbability.shtml Generalized here: http://www.jstor.org/discover/10.2307/2690041?uid=3739616&uid=2129&uid=2&uid=70&uid=4&uid=3739256&sid=55851236913
sorry, last term is 1/r! ^^