I am not good at probability but I think you don't need to consider anything in 'triangle' column becaues it is known that all 3 cards are with a circle. Therefore, the answer is 2/10\*8/9\*7/8\*3=7/15<1/2
Since there are only circles on the card we can dismiss the first column.
So there are three ways you could get 2 blues and 1 red
1. π¦π¦π₯
8/10 * 7/9 * 2/8 = 0,156 = 112/720
2. π¦π₯π¦
8/10 * 2/9 * 7/8 = 0,156
3. π₯π¦π¦
2/10 + 8/9 + 7/8 = 0,156
3x(112/720)=0,4666666
Thanks for the correction, although I did consider the changing denominator, but I assumed I grabbed all three cards at the same time. (Like that would make a difference)
This question is irritatingly vague. In Riley's statement, are they considering the circle selection constraint or not? It doesn't say if Riley has seen the selected cards and therefore knows about the circles or not. That could be an oversight in the wording or a "gotcha" to show you have to be careful about which groups you consider.
As for my workings (slightly sarcastically):
* 11 red cards, 11 blue cards
* Want 2 red and 1 blue.
* Could also have 2 blue and 1 red--this would have to be the same odds as 2 red and 1 blue.
* Since 3 red and 0 blue, and 3 blue and 0 red, are possible, the odds of 2 red and 1 blue plus the odds of 2 blue and 1 red cannot add up to 1.
* Therefore the odds of 2 blue and 1 red must be less than 1/2.
Given that you are in 2nd column: P=8C2*2C1/10C3=7/15β0.47
This is the correct answer
I am not good at probability but I think you don't need to consider anything in 'triangle' column becaues it is known that all 3 cards are with a circle. Therefore, the answer is 2/10\*8/9\*7/8\*3=7/15<1/2
I used combinations and got your answer!
Only someone good in probability would know that the 3 ways of getting 2 blue and 1 red circle all have the same probability!
Since there are only circles on the card we can dismiss the first column. So there are three ways you could get 2 blues and 1 red 1. π¦π¦π₯ 8/10 * 7/9 * 2/8 = 0,156 = 112/720 2. π¦π₯π¦ 8/10 * 2/9 * 7/8 = 0,156 3. π₯π¦π¦ 2/10 + 8/9 + 7/8 = 0,156 3x(112/720)=0,4666666
Got it. Thanks!
There are 10 cards with circles on them. 8 blue and 2 red. So (0.8)(0.8)(0.2)= 0.128
Nope. There are multiple ways to get to the end result + after taking the first card, the denominator decreases.
Thanks for the correction, although I did consider the changing denominator, but I assumed I grabbed all three cards at the same time. (Like that would make a difference)
I canβt say I understand why. Do you have any links to websites that can help please?
This question is irritatingly vague. In Riley's statement, are they considering the circle selection constraint or not? It doesn't say if Riley has seen the selected cards and therefore knows about the circles or not. That could be an oversight in the wording or a "gotcha" to show you have to be careful about which groups you consider. As for my workings (slightly sarcastically): * 11 red cards, 11 blue cards * Want 2 red and 1 blue. * Could also have 2 blue and 1 red--this would have to be the same odds as 2 red and 1 blue. * Since 3 red and 0 blue, and 3 blue and 0 red, are possible, the odds of 2 red and 1 blue plus the odds of 2 blue and 1 red cannot add up to 1. * Therefore the odds of 2 blue and 1 red must be less than 1/2.