the probability of billy will get his own present

Thanks Chris :)

In a party,each of participants,Ada,Billy and 8 of their friends brings along a present for exchange.Each participant will draw a present randomly without displacement.

1)Given that Ada does ont get her own present,find the probability that Billy will get his own present.

I'll give it a shot.  LOL  

There are 10 people

the first person has 1 of 10 possible gifts.

the second person has 1 of 9 possible gifts

In total there are 10! possible combinations.

BUT

I might let Ada go first then Ada will have only 9 not ten gifts to choose from so the number of possibilities will be     9*9!

Now how  many ways can billy get his own present.

I'm going to let billy go first and Ada go second

there is only one way that Billy can get his own gift

then there are 8 ways that Ada can get somone elses

etc

so  maybe that will be      1*8*8!

So maybe 

P(Billy gets his own given that Ada does not get her own) =      $$\frac{8*8!}{9*9!}=\frac{8*1}{9*9}=\frac{8}{91}$$

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I hope that this is not total nonesence because that could be just a tad embarassing.   LOL.

Don't worry - I can take it.  :))    ¯\_(ツ)_/¯

I'll try this one, too

Ada could choose any of 9 gifts and the other 9 people have 9! ways to choose theirs. Note that some of these include the possibiiity that Billy does draw his own gift.

After she draws hers, we want to look at the total scenarios that have Billy "stuck" with his own gift.

In this scenario, Ada draws 8 gifts (since she doesn't draw her own, nor Billy's)....and the other 8 people have 8! ways of choosing amongst the remaining 8 gifts. Notice that some of these include the possibility that they actually draw their own gifts, but that's not a stated consideration.

So  ... P[ Billy draws his own gift l Ada doesn't draw her gift ] = 

[8 * 8! ] /  [9 * 9!] = (8/9)(1/9) = 8/91  ..... the same as found by Melody....