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Mathematics 4 Online
OpenStudy (mathmath333):

Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope

OpenStudy (mathmath333):

Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

OpenStudy (sohailiftikhar):

you are asking about the probability of proper letter for three persons or only one person ? means minimum probability or maximum ?

OpenStudy (mathmath333):

probability that at least one letter is in its proper envelope

OpenStudy (sohailiftikhar):

the probability to get a proper letter for each person is 3/9

OpenStudy (sohailiftikhar):

means 1/3

OpenStudy (mathmath333):

how

OpenStudy (sohailiftikhar):

becz there are three letters and the total probability of these letters to deliver is 9

OpenStudy (sohailiftikhar):

and the probability to get a proper letter for each person is 1/3

OpenStudy (anonymous):

1/3 because that is 3/9 in simplest form

OpenStudy (sohailiftikhar):

get it or not ?

OpenStudy (anonymous):

i get it

OpenStudy (sohailiftikhar):

good

OpenStudy (anonymous):

plus its in simplest form

Parth (parthkohli):

Required probability is given by the number of ways the letters could be placed such that at least one letter is in its correct place divided by the number of ways we could place the letters (favourable ways/total ways). The total number of ways we could keep them is \(3 \times 2 \times 1 = 6\). One out of those ways is when all three are in the correct place. Three other ways when exactly one is in its correct place. Therefore, the number of favourable cases = 4. Now we have P = 4/6 = 2/3.

OpenStudy (mathmath333):

how u got 6

OpenStudy (mathmath333):

The total number of ways we could kee=6

Parth (parthkohli):

It makes sense that the probability should be 2/3 rather than 1/3. It's a lot more probable to keep at least one number in the right envelope.

Parth (parthkohli):

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