Mimi_x3
  • 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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
UnkleRhaukus
  • UnkleRhaukus
What is the probability that one card is the hand is jack of hearts?
Mimi_x3
  • Mimi_x3
1/52
UnkleRhaukus
  • UnkleRhaukus
a hand has 5 cards

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Mimi_x3
  • Mimi_x3
lemme think
Mimi_x3
  • Mimi_x3
it it 1/ 52C5 ?
Mimi_x3
  • Mimi_x3
rawrrr i dont think thats correct
Mimi_x3
  • Mimi_x3
am i completely off?
UnkleRhaukus
  • UnkleRhaukus
5/52
UnkleRhaukus
  • UnkleRhaukus
isn't it?
Mimi_x3
  • Mimi_x3
hmmmm
UnkleRhaukus
  • UnkleRhaukus
so what is the probability of Jack and Queen of hearts?
Mimi_x3
  • Mimi_x3
hmmm i dont think i agree with you
UnkleRhaukus
  • UnkleRhaukus
ok
Mimi_x3
  • Mimi_x3
show me what u got in mind
UnkleRhaukus
  • UnkleRhaukus
1/52 x 1/51 x 1/50 x 39/49 x 38/48 x 5!
UnkleRhaukus
  • UnkleRhaukus
= 5/52 x 4/51 x 3/50 x 2*39/49 x 38/48
Mimi_x3
  • Mimi_x3
Interesting .... This is what I got Total sample space is 52C5 right?
Mimi_x3
  • Mimi_x3
like we have 52 cards and we only choose 5
UnkleRhaukus
  • UnkleRhaukus
yeah
Mimi_x3
  • 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)
UnkleRhaukus
  • UnkleRhaukus
Our results agree.
Mimi_x3
  • Mimi_x3
oh yea???
UnkleRhaukus
  • UnkleRhaukus
19/ 33320 = 0.0000570....
Mimi_x3
  • Mimi_x3
sweeet
Mimi_x3
  • 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
Mimi_x3
  • Mimi_x3
Why you multiplying by 5! at the end?
UnkleRhaukus
  • UnkleRhaukus
because the 5 cards can be in any order
Mimi_x3
  • Mimi_x3
Oh ok i Get what you did .... Thanks
Mimi_x3
  • Mimi_x3
k I got one more question
Mimi_x3
  • 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
UnkleRhaukus
  • UnkleRhaukus
6/13 x 5/12 x ...
Mimi_x3
  • Mimi_x3
Im just wondering if I was to use my method what would be the total sample space
UnkleRhaukus
  • UnkleRhaukus
13 C 5?
Mimi_x3
  • Mimi_x3
how abt the girl and boy thing and divided into groups
Mimi_x3
  • Mimi_x3
ughhhh ur right -.-
Mimi_x3
  • Mimi_x3
dang it it was sooo simple
Mimi_x3
  • 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?
UnkleRhaukus
  • UnkleRhaukus
that will work
Mimi_x3
  • Mimi_x3
how were you thinking of approaching this problem?
UnkleRhaukus
  • UnkleRhaukus
there are only the three cases to consider, so this is probably the easiest way
Mimi_x3
  • Mimi_x3
Ok Thanks!!!!!!!!!! I really appreciate it that you took the time to help me

Looking for something else?

Not the answer you are looking for? Search for more explanations.