View Full Version : 5 checks randomly inserted into addressed envelopes....

KingAce

03-04-2008, 10:35 PM

i'm having some serious difficulty with this question and would really appreciate anyone that could even get the answer (i really think its impossible haha).

The problem is as follows: Five paychecks and envelopes are addressed to five different people. The paychecks are randomly inserted into the envelopes. What is the probability that (a) exactly one paycheck is inserted in the correct envelope and (b) at least one paycheck is inserted in the correct envelope?

The answer for (I believe) part B is 45/120.

I just don't know how to get it.

Thanks for your help! :D

Denis

03-04-2008, 11:49 PM

http://mathforum.org/library/drmath/view/56592.html

soroban

03-05-2008, 02:32 AM

Hello, KingAce!

Five paychecks and envelopes are addressed to five different people.

The paychecks are randomly inserted into the envelopes.

What is the probability that:

(a) exactly one paycheck is inserted in the correct envelope?

(b) at least one paycheck is inserted in the correct envelope?

There are:.5! \:=\;120 possible placements of the paychecks.

(a) One paycheck is its proper envelope.

There are 5 choices for this paycheck.

The other four paychecks, A,B,C,D are not in thieir respective envelopes.

There are 9 ways for this to happen. .*

Hence, there are: .5\cdot9 \:=\:45 ways for exactly one paycheck to be correct.

\text{Therefore: }\;P(\text{1 correct}) \:=\:\frac{45}{120} \:=\:\frac{3}{8}

(b) The opposite of "at least one correct" is "none correct".

For all five paychecks to be in the wrong envelopes, there are 44 ways. .**

Hence, there are: .120 -44 \:=\:76 ways for at least one paycheck to be correct.

\text{Therefore: }\;P(\text{at least 1 correct}) \;=\;\frac{76}{120} \;=\;\frac{19}{24}

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

*

If the proper order is ABCD, there are 9 "disarrangements".

. .\begin{array}{ccc}BADC &CADB& DABC \\ BDAC & CDAB & DCAB \\ BCDA & CDBA & DCBA\end{array}

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

There is bizarre formula for the number of disarrangements of n objects.

Example: n = 5

Expand (Q - 1)^5

. .We have: . Q^5 - 5Q^4 +10Q^3- 10Q^2 + 5Q - 1

\text{Let }\,Q^n \:=\:n! . . . (Imagine! .An exponent becomes a factorial!)

. . We have: . 5! -5\!\cdot\!4! + 10\!\cdot\!3! -10\!\cdot\!2!+ 5\1\cdot\!1! - 1 \;=\;120 - 120 +60-20 + 5 - 1 \;=\;44

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