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Probability

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Re: Probability

by logitech » Tue Nov 18, 2008 9:16 am
moneyman wrote:Guys pls explain this one
4 letters, 4 envelops

L1,L2,L3,L4

E1,E2,E3,E4

so can have 4x3x2x1 = 24 total different ways

Lets say, L1 is in E1

L1 L3 L4 L2
E1 E2 E3 E4

L1 L4 L2 L3
E1 E2 E3 E4

Only two ways in which the other three letters are in wrong envelops

since we have 4 letters , 4 x2 = 8 ways

8 / 24 = 1/3
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by rtfact » Tue Nov 18, 2008 9:39 am
suppose we have

envelopes: a,b,c,d
addresses: 1,2,3,4

suppose a has to be sent to 1 (1 option),
AND b has to be sent to 3 or 4 (2 options). suppose b is sent to 3.
AND c has to be sent to 2 or 4 (2 options). suppose c is sent to 2.
AND d is sent to 4. (1 option).

all said above was under the assumption that 'a' was sent to the correct address, but it could be equally be 'b','c', or 'd'. so (4 options).

multiply 1*2*2*1*4=8
8 out of 4!=8/24=1/3.
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