An urn contains 10 red balls, 10 green balls, and 5 white balls. 5 balls are selected. In how many ways can 5 balls be drawn if at least 3 are green?

Question

anonymous · Answer

Someone please help!!!!

loser66 · Answer

@ganeshie8  I love the way you break the green as an independent variable. Please, do it again. :)

anonymous · Answer

can someone just help me please?

mathmath333 · Answer

@JosephDeng  what is the answer given in ur book

mathmath333 · Answer

i think it is $\dbinom{15}{2}+\dbinom{15}{1}+1=121$

mathmate · Answer

@JosephDeng 
Hint:
We just need to know that there are 10 green and 15 non-green.
The number of combinations of pulling 3 green (10 choose 3)  and 2 non green (15 choose 2)  =$\dbinom{10}{3}\dbinom{15}{2}=12600$
Total number of combinations (25 choose 5) = $\dbinom{25}{5}=53130$
Can you take it from here to find P(at least 3 green)?

anonymous · Answer

the answer is 16002

mathmate · Answer

Yes, the answer is correct.
$\dbinom{10}{3}\dbinom{15}{2}+\dbinom{10}{4}\dbinom{15}{1}+\dbinom{10}{5}\dbinom{15}{0}=16002$