Probability question (1 Viewer)

[*girl04*]

Hi could anyone tell me how to approach this problem ( please). I tried a tree-diagram but it got messy.
question
A game of poker uses a deck of 52 cards with 4 suits. Each suit has 4 cards ( ace, 2-10, queen, king and jack) If a person is dealt 5 cards find the probability of getting
* 4 aces
* a flush ( all cards of the same suit)
Btw the answers are 1/54145 and 33/16660 thanks!

 

[Jago]

each suit has 4 cards??? nevertheless...

4 aces - (4/52) x (3/51) x (2/50) x (1/49) x 5

flush - 1 x (12/51) x (11/50) x (10/49) x (9/48)

 

[*girl04*]

each suit has 4 cards??? nevertheless...

4 aces - (4/52) x (3/51) x (2/50) x (1/49) x 5

flush - 1 x (12/51) x (11/50) x (10/49) x (9/48)

thanks for your quick response!
With the 1st answer why did you times it by 5? and not do that with the 2nd one?
thanks

 

[Jago]

i did it becuase of 5C4 (3u method), but im not too sure how to explain it to you. Maybe someone else can give you a more literate response, but in 1 you need 4 distinct cards whereas in 2 you needed 5.

 

[rukie]

ok i think this is an easier method,

so the combination has to be A A A A ?, coz 4 cards are used ? has 48 possibilities
so there the possibilities to get 4 aces, just divide that by total number of possiblities (52C5). think that gives you the right answer

48/52C5 = 1/54145

 

[speed2]

so clever!!

 

[nit]

Similarly for the last question, the easy method is p= 4*(13C5)/52C5

Each suit comprises 13 cards, from which you desire 5, and there are 4 suits. The probability is given by dividing through by the total number of choices of 5 cards from 52.

 

[haboozin]

Similarly for the last question, the easy method is p= 4*(13C5)/52C5

no C or P in 2unit...

 

[word.]

The 4-aces is a 3u probability question and aren't asked in the mathematics exam..


You can always work towards the binomial probability taught in 3u from mathematics level anyway:

A = ace, O = other
you can either get AAAAO, AAAOA, AAOAA, AOAAA and OAAAA...

so with 5 combinations:

P(4 aces) = {(4/52) * (3/51) * (2/50) * (1/49) * (48/48)} + {(4/52) * (3/51) * (2/50) * (48/49) * (1/48)} + ...

at this point, better 2u students should see the pattern:
the probability of getting 4 aces in the 1st order is the same as the 2nd..
and as there are 5 orders it can be achieved, just take the first order and multiply by 5.

either that or they can just keep going for all the orders

 

[*girl04*]

^ really? coz i do 2 unit maths and it was in our textbook

 

[DraconisV]

Is all of this probablity with this C5 thingies HSC work, coz i didnt do probability as a topic in 2 unit or ext 1 in the prelim

 

[acmilan]

Is all of this probablity with this C5 thingies HSC work, coz i didnt do probability as a topic in 2 unit or ext 1 in the prelim

I remember discussing this a while ago. I did it in the prelim but according to everyone else its a HSC topic.

 

[Jago]

Permutations and Combinations is HSC Extension

 

[DraconisV]

Thew, thx Jago

 

[StGeorgeDragons]

ok i think this is an easier method,

so the combination has to be A A A A ?, coz 4 cards are used ? has 48 possibilities
so there the possibilities to get 4 aces, just divide that by total number of possiblities (52C5). think that gives you the right answer

48/52C5 = 1/54145

how is there 48 possbilities? from AAAA

 

[speed2]

how is there 48 possbilities? from AAAA

i think this is what he means; there are 5 cards given out right, 4 of them are taken up by A, i.e. AAAA_ the _ can be any of the remainig 48 cards.

 

[StGeorgeDragons]

each suit has 4 cards??? nevertheless...

4 aces - (4/52) x (3/51) x (2/50) x (1/49) x 5

flush - 1 x (12/51) x (11/50) x (10/49) x (9/48)

i dont understand why it was multiplied by 5?

 

[karen88]

its x 5 cos it has 5 combinations eg AAAAO AAAOA AAOAA AOAAA OAAAA