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Suppose each of three men at a  party throws his hat into the center of the room. The hats are first mixed up and each man randomly selects a hat. what is the probability that-

a. All of three men select his own hat.

b.None of three men select his own hats.

### 1 comment

Basic Formula :-

No. of De-Arrangement with n = $n!\; [ \; \frac{1}{0!}\; - \; \frac{1}{1!}\; + \; \frac{1}{2!}\; - \; \frac{1}{3!}\; + .... +\; \frac{(-1)^n}{n!}\; ]$

Given that n=3, ==> No.of de-arrangements = 3! * [$\frac{1}{2}\;-\frac{1}{6}$] = 3! * [$\frac{1}{3}$]

Sample space with selecting n hats by n persons =  n! = 3! =6

1) All of them Selecting their own Hats = only one way.

∴ P( All of three men select his own hat. ) = $\frac{1}{n!}$ =$\frac{1}{3!}$ = $\frac{1}{6}$

2) None of them Selecting their own Hats = No.of De-arrangements

∴ P( None of three men select his own hat. ) = $\frac{n!\; [ \; \frac{1}{0!}\; - \; \frac{1}{1!}\; + \; \frac{1}{2!}\; - \; \frac{1}{3!}\; + .... +\; \frac{(-1)^n}{n!}\; ]}{n!}$ = $\; [ \; \frac{1}{0!}\; - \; \frac{1}{1!}\; + \; \frac{1}{2!}\; - \; \frac{1}{3!}\; + .... +\; \frac{(-1)^n}{n!}\; ]$ = [$\frac{1}{3}$]

(3) What is the probability that exactly 1 men gets their own hat ?

if there are k mens gets their original hat, then there should be n-k mens should not get their original hats.

===> n = ( k  original hats = 1 way only. ) + ( n-k dearrangements )

n-k dearrangements = $p!\; [ \; \frac{1}{0!}\; - \; \frac{1}{1!}\; + \; \frac{1}{2!}\; - \; \frac{1}{3!}\; + .... +\; \frac{(-1)^p}{p!}\; ]$, where p = n-k.

according to the question, p=3-1 = 2 ===> p dearrangements = 2 *[ $\frac{1}{2}$] = 1

Exactly K mens get their OWN hats = $\binom{n}{n-k}$ * (n-k dearrangements)

∴ P( Exactly K mens get their OWN hats ) =$\binom{n}{p}$ $\frac{p!\; [ \; \frac{1}{0!}\; - \; \frac{1}{1!}\; + \; \frac{1}{2!}\; - \; \frac{1}{3!}\; + .... +\; \frac{(-1)^p}{p!}\; ]}{n!}$ = $\binom{3}{2}$ * $\frac{1}{3!}$ = 3 * $\frac{1}{3!}$ =$\frac{1}{2!}$ = $\frac{1}{2}$

reshown

Answer of second is given 2/3.

it should be wrong... just check it manually, there are 3! ways

i) 1,2,3

ii) 1,3,2

iii) 2,3,1

iv) 2,1,3

v) 3,1,2

vi) 3,2,1

in this 6 orders, only two orders i.e., iii and v are favorable cases ===>probability =  $\frac{2}{6}$=  $\frac{1}{3}$
Yes, you are right. The answer is correct.

Thanks.