General DiscussionMath geek HELP!

10% you get 2 orange ones

2.38% for the first one.
14.28% for the second.

op is 13 years old EleGiggle

Tell me how u calculate please

And isnt both answer the same? 6/36 x 5/35 for both?

That's the calc for the first one. But in the second case you already know the first is red. So it's only 5/35

Huh but u pick the first orange isnt it 6/36?

If you pick the first orange out and put it aside it takes one candy out of the basket, making there be 5 oranges and 30 yellows, 30+5=35.
Therefore the probability of you getting orange again is 5/35, or 1/7.

(6/36)*(5/35)
and i'm not sure how is the other one meant. either it's the same or it's 5/35 since you're choosing only from 5 orange in 35 sweets

unless i'm missing on something.. i'm not really good at combinatorics, this stuff is quite boring.

i think ur chance is about 99.756%, by the way

its the same thing. It will be 6/36 * 5/35. Doesn't matter if they're picked together or simultaneously

That's what i thought @undertale

C(6,2)/C(36,2)

around 0,5 % chance to be blind

What YAN did is technically the correct way of doing it, but the value in the end is the same anyway.

If you look at the first one and see it is orange. Then the chance if the second one also being orenge is 5/35.
Not sure if that's what you meant.

if you're a space cow, 100%.

If you put the first one back before picking the second, it's 1/36 chance overall. If you leave the first one out before picking the second, it's 1/42 chance.

The possibility of ur goal is (number of oranges/total) . U want "only and only 2 oranges instant",when u go.So the possibilities are multiplied.
p1*p2.

1st case and 2nd are the same in fact. p1= 6/36 * p2=5/35;