Question

There are 10 red balls 20 blue balls and 30 green balls in a bag and 5 balls are Selected from it find the probability that all with be blue ? (b) Atleast one will be green

Answer 1

P(all will be blue) =20C5 / 60C5

Answer 2

Yes! Good work!

Answer 3

P(Atleast one green) = 20C1/60C5 * 20C2/60C5 * 20C3/60C5 * 20C4/60C5 * 20C5/60C5 *

Answer 4

For the second part, instead of adding up P(1) + P(2) + ... + P(5) for at least one green. It is better to get P(no green balls) and subtract that from 1.

Answer 5

So, 30 "non-greens" in 60.

Answer 6

So, 1 - 30C5/60C5

Answer 7

\[1-P(\text{0 are green})=1-(\frac{30}{60}\frac{29}{60}\frac{28}{60}\frac{27}{60}\frac{26}{60}))\]

Answer 8

Assuming balls are not replaced

Answer 9

@henpen , that's not right because that formula numerator on the right-side term is going to give you permutations instead of the combinations that you want.

Answer 10

I think you're overcomplicating.

Answer 11

Sorry- I made an \[1-(\frac{30}{60}\frac{29}{59}\frac{28}{58}\frac{27}{57}\frac{26}{56})\]error earlier

Answer 12

Formula is clear and in its most simplified form and given already. You can't get simpler than what I already gave: So, 1 - 30C5/60C5

Answer 13

I am answering b, by the way

Answer 14

Yes, so am I, but you are not answering anywhere near correctly.

Answer 15

Why evoke combinatorics? There is only 1 way for there not to be at least 1 green.

Answer 16

That is, GGGGG

Answer 17

\[P(\text{at least 1 green})=1-P(G,G,G,G,G)\]Do you agree with this?

Answer 18

%15 percent will Blue.

Answer 19

@henpen ,Your corrected formula will work and can be derived from what I already gave. If there were a denominator of 5! for both numerator and denominator, then they are the same. Now that you corrected your formula, you are on the right track. Same answer. The reason I put the answer in that form is because that is how the soultion is classically represented and is the most straightforward logically-speaking.

Answer 20

I'm sorry for the error earlier. It just seemed silly to haul in nCr notation where intuition suffices.

Answer 21

Not, silly. Logical. For those not familiar with the mathematical theory, it is good to reinforce the theory, again, your formula is derivable from mine. It is the same, so no quibbling is necessary here.

And the numerical answer for part #2 is then approx. 0.9739

Answer 22

thxxx guys

Answer 23

yw!