I got these very easy prob. questions but none of my answers match those of the book except for the 1st and last one. Here's the problem,(adsbygoogle = window.adsbygoogle || []).push({});

Dice poker is played by throwing simultaneously 5 dices. Show that:

a) P(5 different cards)=0.0926

b) P(1 pair)=0.4630

c) P(2 pairs)=0.2315

d) P(3 of a kind)=0.1543

e) P(full house)=0.0386

f) P(4 of a kind)=0.0193

g) P(5 of a kind)=0.0008

Solution:

The fundamental set [itex]\Omega[/itex] is "all possible thrown of the 5 dices", which yields #[itex]\Omega=6^5[/itex]. We make the hypothesis that all fundamental events (the elements of [itex]\Omega[/itex]) are equiprobable. With this assumption, we can calculate the required probability by taking the ratio of the given event to the cardinality of [itex]\Omega[/itex].

a) P(5 different cards)=6*5*4*3*2/6^5 = 0.0926

b) There are 6 ways of chosing the pair, then 5*4*3 ways to choose the other 3 cards so they are not equal to one another or to the pair. Hence,

P(1 pair)=6*5*4*3/6^5=0.0463.

c) P(2 pairs)= 6*5*4/6^5=0.0154

d) P(3 of a kind)=6*5*4/6^5=0.0154

e) P(full house)=6*5/6^5=0.00386

f) P(4 of a kind)=6*5/6^5=0.00386

g) P(5 of a kind)=6/6^5=0.0008

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# Homework Help: Very quick probability verification

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