Question

find the probability the first three cards chosen are a queen and 2 diamonds.

Answer 1

I need help with stats

Answer 2

so there's 52 cards in a deck, 4 queens in all(1 in each club) and i think 13 diamonds (1-10, the king and queen, and the jack), so take the combined probability of the queens and diamonds 17/52 and multiply it by 3/52

Answer 3

So do I multiply since its "and"

Answer 4

multiply the combined probability of the total queens and diamonds in a deck by the probability of the cards at hand which is 3 out of 52.

Answer 5

Oh alright, thank you very much

Answer 6

and with that answer too(i forgot 2/3 were diamonds) multiply the ratio of two stacks of diamonds which is 26/52=1/2 by that previous result(17/52*3/52) and do the same for 1 queen 1/52

Answer 7

oh that's what was missing

Answer 8

yeah i had to clean that up ;)

Answer 9

thx buddy

Answer 10

If we assume that the chosen queen can not be the queen of diamonds, the queen of diamonds needs to excluded from the choice of queen (giving 3 queens to choose from) and also excluded from the choice of the 2 diamonds (avoiding getting 2 queens).
The required probability is therefore given by:
\[\large P(1\ queen,\ 2\ diamonds)=\frac{3C1\times 12C2}{52C3}=\frac{3\times12\times11\times3\times2}{2\times52\times51\times50}=you\ can\ calculate\]

Answer 11

hmmm, didn't know i had to use the combination rule

Answer 12

thanks

Answer 13

You're welcome. The problem is sampling without replacement of the multivariate type.
"Find the probability the first three drinks chosen are diet sprite and 2 waters (in that order)"
Please post as a new question. Note:The populations of the types of drink needs to be included to make a solution possible.

Answer 14

ok

Answer 15

sorry for that