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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.

3/9=1/3

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the answer is not 1/3 its 2/3

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it's 1/3

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 2/3

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 .We need to find the probability that atleast one letter was put in the correct envelope.

The total number of ways of putting three letters in 3 envelopes = 3 x 2 x 1 = 6
The number of ways of putting all three letters in incorrect envelope is =2
Letter for X goes into envelope for Y and Letter for Y goes into envelope for Z, Letter for Z goes
 into envelope for X 
or Letter for X goes into envelope for Z and Letter for Y goes into envelope for X and Letter for Z goes into
envelope for Y ]
2 1
P(None of the letters is put in the right envelope)  = 
6 3
P(Atleast one letter is put in the right envelope)=
=
1 2
 1 _ - P(None of the letters is put in the right envelope)  1
3 3
 
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guys for the same question what if all the letters go into the right envelopes
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Total ways=3!=6 At least one current=exactly one in correct + all in correct=1+3=4 So P(at least one in correct) =4/6=2/3
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What......
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?We need to find the probability that atleast one letter was put in the correct envelope.

The total number of ways of putting three letters in 3 envelopes = 3 x 2 x 1 = 6

The number of ways of putting all three letters in incorrect envelope is =2

Letter for X goes into envelope for Y and Letter for Y goes into envelope for Z, Letter for Z goes

?into envelope for X?

or Letter for X goes into envelope for Z and Letter for Y goes into envelope for X and Letter for Z goes into

envelope for Y ]

2 1

P(None of the letters is put in the right envelope) ?=?

6 3

P(Atleast one letter is put in the right envelope)=

=

1 2

?1 _ - P(None of the letters is put in the right envelope) ?1

3 3.
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