3,728 views
0 votes
0 votes

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 Answer

Best answer
3 votes
3 votes

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 * $\frac{1}{3!}$ =$\frac{1}{2!}$ = $\frac{1}{2}$

selected by

Related questions

0 votes
0 votes
1 answer
1
Pooja Khatri asked Sep 25, 2018
265 views
Let $X$ be a random variable with $P(X=1) =0.2,P(X=2)=0.3$, and $P(X=3)=0.5$. What is the expected value of $X$?
1 votes
1 votes
2 answers
3