Re: Ram had to post 5 letters to 5 different persons. He put the l
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Updated on: 08 Dec 2019, 21:48
Let the 5 letters be L1, L2, L3, L4 and L5 and their correct envelopes be E1, E2, E3, E4 and E5
For favourable outcomes:
Select any two letters.
Two letters out of 5 can be selected in 5C2 ways.
The selected 2 letters can be put in their respective envelopes in only 1 way i.e. 5C2 × 1 = 5C2 ways. (e.g. L2 and L3 were selected, they can be put in E2 and E3 in only 1 way.)
But to get exactly 2 letters correctly posted, the remaining 3 letters must be put in the incorrect envelopes.
Let the remaining letters and envelopes be L1, L4, L5 and E1, E4, E5.
Correct combination is
L1 - E1
L4 - E4
L5 - E5
Possible incorrect combinations are
1. L1-E4, L4-E5, L5-E1.
2. L1-E5, L4-E1, L5-E4.
Therefore, total number of favourable outcomes will be
combination of correct letters * combination of incorrect letters = 5C2 * 2
For total number of outcomes:
L1 can be put in any of the 5 envelopes. Therefore, 4 envelopes will be left for 4 letters.
Now, L2 can be put in any of the remaining 4 envelopes. Therefore, 3 envelopes will be left for 3 letters.
Similarly, L3, L4 and L5 can be put in any of the remaining 3, 2 and 1 envelopes respectively.
Therefore, total number of outcomes = 5×4×3×2×1 = 5!
Therefore, probability, P = favouravle outcomes/ total no. of outcomes = (5C2 * 2)/5! = 20/120 = 1/6
Posted from my mobile device
Originally posted by
jhavyom on 08 Dec 2019, 04:46.
Last edited by
jhavyom on 08 Dec 2019, 21:48, edited 1 time in total.