(non-3U probability actually) (1 Viewer)

[mojako]

Can somebody explain the riddle in http://www.boredofstudies.org/community/showthread.php?t=43995
??

Also I came across a similar-looking question from Cambridge, Extension section.
In a game, a player draws a card from a pack of 52. If he draws a two he wins. If he draws a three, four or five he loses. If he draws a heart that is not a two, three, four or five then he must roll a die. He wins only if he rolls a one. If he draws one of the other three suits and the card is not a two, three, four or five, then he must toss a coin. He wins only if he tosses a tail. Given that the player wins the game, what is the probability that he dew a black card?
Answer is 11/19.
How do we do this one???

Thanks.

 

[withoutaface]

Ok so you first have the winning by drawing a 2 that is read, and the chance is 1/26. Then you have the chance of winning by non 2 hearts which is 9/52*1/6, and the prob of winning with a non 2 diamond is 9/52*1/2

Then you have the prob of winning by a black 2 which is 1/26. The prob of winning by a black non 2 is 9/52*1/2+9/52*1/2

then you have (win by black)/(win by red+win by black)=11/19

 

[Estel]

If he won, the only appropriate sample space is;
draw a 2, draw a non-2,3,4,5 hearts then roll 1, draw a non-2,3,4,5 non hearts then roll a coin.

Respectively these probabilities are: 1/13, 9/52*1/6, 27/52*1/2
ie 1/13, 3/104, 27/104
Relatively these probabilities are: 4/19, 3/38, 27/38 (i.e. the probabilities of each if we know one of these three outcomes happened)

Probability of drawing a black card for each outcome is 1/2, 0, 2/3 respectively.
Hence required probability is 1/2*4/19+2/3*27/38 = 11/19

meh beaten to it. I'll leave this here for the benefit of a 2nd explanation.

 

[withoutaface]

If he won, the only appropriate sample space is;
draw a 2, draw a non-2,3,4,5 hearts then roll 1, draw a non-2,3,4,5 non hearts then roll a coin.

Respectively these probabilities are: 1/13, 9/52*1/6, 27/52*1/2
ie 1/13, 3/104, 27/104
Relatively these probabilities are: 4/19, 3/38, 27/38 (i.e. the probabilities of each if we know one of these three outcomes happened)

Probability of drawing a black card for each outcome is 1/2, 0, 2/3 respectively.
Hence required probability is 1/2*4/19+2/3*27/38 = 11/19

meh beaten to it. I'll leave this here for the benefit of a 2nd explanation.

Somebody just wasted 15 mins

 

[Estel]

Well not really, I've been meaning to type up the complete solutions to cambridge 3U at some stage


edit: bah I'm bitter, could've just used ur solution =p

 

[mojako]

Oh thank you !

@withoutaface:
Can you explain the "(win by black)/(win by red+win by black)=11/19" part?

@Estel:
How did you find the relative probabilities?

 

[withoutaface]

well your assuming he's already won, so hence your just finding the probability of winning by black compared to that of winning by red. So the answer is essentially (win by black)/(win)

 

[Estel]

mojako: relative probabilities was a term I shouldn't have used, it's quite incorrect.

What I mean is the probability of each given that one of the 3 has occured (i.e. the explanation I gave)
Suppose that I have a bag of marbles, P(red) = 1/4, P(blue) = 1/3, P(green) = 5/12 (pray that adds to 1 ) and I know green is definitely not the marble I selected, then
P(red) = (1/4)(1/4+1/3) and P(blue) = (1/3)/(1/4+1/3)
I hope this analogy helps you whilst giving something to think about.