Permutation question (final exam tomorrow)

[DenX2]

Hi i came across this practice question that i have no idea on how to go about it.

 

Five balls are randomly drawn without replacement from a bag containing 4 red balls and 6 black balls. What is the probability that at least 3 red balls will be drawn?

the answer is 0.2619

 

I've tried tree diagrams and the formulas but none of them work, may someone please teach me how???

[greentea]

note: i use . for multiplication

 

what is the number of possible combinations (or whatever this is called, i never remembered them) of five balls out of 10?

10.9.8.7.6/(5.4.3.2.1) = 252

why?

for the first one you have 10 choices, for the second 9, etc (since they are not replaced)

you should also take into account that you do not care about the order. only which balls are taken. since there are 5.4.3.2.1 ways to order five balls divide by that number

example: you took balls 9 4 2 1 7 or 7 4 1 2 9 - same for you. so 120 ways to rearrange those.

that is true for each set of five balls you took out of the ten. therefore divide to get the 'different' combinations (in the sense we are interested).

next, let's look at color

to get at least 3 balls you need either 3 or 4 (all)

in a similar manner you can determine the combinations 4.3.2/(3.2.1) for the 3 red balls. then you must also have 2 black balls that make 6.5/(2.1) combinations. Total = 4.15=60

in the case of 4 red balls 4.3.2.1/(4.3.2.1) - only one combination in which you have them all, kind of obvious. for the remaining black 6/1=6. Total = 1.6=6

Total combinations for 3 or 4 red balls 60+6=66

66/252=0,2619

I am explaining in a quite poor way since i do not remember what the proper terms were.