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Mathematics 18 Online
OpenStudy (mimi_x3):

A poker hand of 5 cards is dealt from a standard deck. What is the probability that the jack, queen, and king of hearts but no other hearts are in the same hand?

OpenStudy (unklerhaukus):

What is the probability that one card is the hand is jack of hearts?

OpenStudy (mimi_x3):

1/52

OpenStudy (unklerhaukus):

a hand has 5 cards

OpenStudy (mimi_x3):

lemme think

OpenStudy (mimi_x3):

it it 1/ 52C5 ?

OpenStudy (mimi_x3):

rawrrr i dont think thats correct

OpenStudy (mimi_x3):

am i completely off?

OpenStudy (unklerhaukus):

5/52

OpenStudy (unklerhaukus):

isn't it?

OpenStudy (mimi_x3):

hmmmm

OpenStudy (unklerhaukus):

so what is the probability of Jack and Queen of hearts?

OpenStudy (mimi_x3):

hmmm i dont think i agree with you

OpenStudy (unklerhaukus):

ok

OpenStudy (mimi_x3):

show me what u got in mind

OpenStudy (unklerhaukus):

1/52 x 1/51 x 1/50 x 39/49 x 38/48 x 5!

OpenStudy (unklerhaukus):

= 5/52 x 4/51 x 3/50 x 2*39/49 x 38/48

OpenStudy (mimi_x3):

Interesting .... This is what I got Total sample space is 52C5 right?

OpenStudy (mimi_x3):

like we have 52 cards and we only choose 5

OpenStudy (unklerhaukus):

yeah

OpenStudy (mimi_x3):

Event A = Picking Jack of hearts, King of hearts and Queen of hearts and 2 other cards that are not a heart Event A = 1*1*1*(52-13)*(52-14) Event A = 1*1*1*39*38 Probability of Event A = Event A /Total Sample Space =(39*38)/(52C5)

OpenStudy (unklerhaukus):

Our results agree.

OpenStudy (mimi_x3):

oh yea???

OpenStudy (unklerhaukus):

19/ 33320 = 0.0000570....

OpenStudy (mimi_x3):

sweeet

OpenStudy (mimi_x3):

I guess in that case i can assume that my method is correct Just for curiosity sake I would love to understand your method

OpenStudy (mimi_x3):

Why you multiplying by 5! at the end?

OpenStudy (unklerhaukus):

because the 5 cards can be in any order

OpenStudy (mimi_x3):

Oh ok i Get what you did .... Thanks

OpenStudy (mimi_x3):

k I got one more question

OpenStudy (mimi_x3):

Five students are randomly selected from seven boys and six girls to join a ski trip. What is the probability that All are girls

OpenStudy (unklerhaukus):

6/13 x 5/12 x ...

OpenStudy (mimi_x3):

Im just wondering if I was to use my method what would be the total sample space

OpenStudy (unklerhaukus):

13 C 5?

OpenStudy (mimi_x3):

how abt the girl and boy thing and divided into groups

OpenStudy (mimi_x3):

ughhhh ur right -.-

OpenStudy (mimi_x3):

dang it it was sooo simple

OpenStudy (mimi_x3):

Gonna drive you insane one last time butttttt What is the probability if there were more boys than girls In that case we gotta look at 3 options : 5 Boys 0 Girls, 4 Boys and 1 Girl , 3 Boys and 2 Girls So in this event you would take the probability of each scenario and then sum them all up correct?

OpenStudy (unklerhaukus):

that will work

OpenStudy (mimi_x3):

how were you thinking of approaching this problem?

OpenStudy (unklerhaukus):

there are only the three cases to consider, so this is probably the easiest way

OpenStudy (mimi_x3):

Ok Thanks!!!!!!!!!! I really appreciate it that you took the time to help me

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