anonymous
  • anonymous
From a deck of 52 cards, how many 5-card hands can be formed with at least 3 diamonds?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
there are 13 diamonds divided by 3 gives 4 possible decks with at least 3 diamonds
anonymous
  • anonymous
wut.
Zarkon
  • Zarkon
A lot. What have you tried?

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More answers

anonymous
  • anonymous
im not sure, i feel like it starts off with \[? nCr 5 \]?
Zarkon
  • Zarkon
do you know how many hands if there are exactly 3 diamonds?
anonymous
  • anonymous
:/ no...
anonymous
  • anonymous
5 hand?
Zarkon
  • Zarkon
how many ways can you choose 3 cards from 13?
anonymous
  • anonymous
so 13 Cr 3?
anonymous
  • anonymous
o wait
anonymous
  • anonymous
i meant 13 nCr 3
Zarkon
  • Zarkon
drop the r \[\Large_{13}C_3\]
Zarkon
  • Zarkon
ok how many cards are not diamonds?
anonymous
  • anonymous
i don't know
Zarkon
  • Zarkon
a card is a diamond or it is not. you have 52 total cards and 13 of them are diamonds
anonymous
  • anonymous
39
Zarkon
  • Zarkon
ok
Zarkon
  • Zarkon
how many ways can you choose the remaining 2 card from the 39?
anonymous
  • anonymous
38?
anonymous
  • anonymous
so 39 * 38 * 37
Zarkon
  • Zarkon
no
Zarkon
  • Zarkon
this is again a combination
Zarkon
  • Zarkon
you are CHOOSING 2 cards from 39
anonymous
  • anonymous
what where did the 2 come from
Zarkon
  • Zarkon
you are picking a 5 card hand. 3 are diamonds...thus the other 2 cards must not be diamonds
anonymous
  • anonymous
oh psh right :/
anonymous
  • anonymous
um so after 2, 37's left
Zarkon
  • Zarkon
I don't care how many are left after you pick 2. We need to know how many ways we can pick the 2 cards from the 39
Zarkon
  • Zarkon
it is the same exact idea we used to choose the 3 diamonds
anonymous
  • anonymous
39 C 2
anonymous
  • anonymous
and then multiply it with 13 C 3
Zarkon
  • Zarkon
yes
Zarkon
  • Zarkon
but that is not the final answer
Zarkon
  • Zarkon
because they are asking for AT LEAST 3 diamonds we need to do this tow more times. one with 4 diamonds and one with 5 diamonds
Zarkon
  • Zarkon
*two
Zarkon
  • Zarkon
N(at least 3 diamonds) =N(get exactly 3 diamonds)+N(get exactly 4 diamonds)+N(get exactly 5 diamonds)
anonymous
  • anonymous
13 C 4 * 38 C 1; 13 C 5 * 37 C 0
anonymous
  • anonymous
oh
Zarkon
  • Zarkon
almost
Zarkon
  • Zarkon
there are always 39 cards that are not diamonds
anonymous
  • anonymous
LOL shet
anonymous
  • anonymous
ok
anonymous
  • anonymous
so it's 13 C 4 * 39 C 1; 13 C 5 * 39 C 0
Zarkon
  • Zarkon
yes...so what is the final answer?
anonymous
  • anonymous
211926?
Zarkon
  • Zarkon
no
anonymous
  • anonymous
oh no
anonymous
  • anonymous
crap
anonymous
  • anonymous
one sec
anonymous
  • anonymous
241098
Zarkon
  • Zarkon
yes
anonymous
  • anonymous
LOL THANK YOU SO MUCH
Zarkon
  • Zarkon
no problem

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