very hard math riddles

[basenji7 1,199]

For the first question - 5 apples and 3 oranges again. We both ate from the furits I gave her.

Ok, so I have 5 apples and 3 oranges. There is a 5/8 chance that I ate an apple and a 3/8 change that I ate an orange. Then I bought 5 apples and 3 oranges again. At this point, the probability of having 10 apples and 5 oranges is 1/2 and the probability of having 9 apples and 6 oranges is 1/2. 

 

 

Let's start with Case 1, 10 apples and 5 oranges. There are four possibilities of the fruits I gave Annie, three apples, two, one, or none. The total amount of ways to choose the fruits I give Annie is 15 choose 3, which is 455. 

 

Case 1.1, three apples. The number of ways that this can happen is 10 choose 3, which is equal to 120. So the probability for this occurring 120/455 = 24/91. After this, the probability that I get an orange and she gets an apple is 0, since there are no oranges. 

 

Case 1.2, two apples and an orange. The number of ways that this can happen is 10 choose 2 times 5 = 225, since we're choosing 2 apples and an orange. So the probability that this will happen is 45/91. The probability that I get an orange is 1/3 and the probability that she gets an apple after that is 1. Then the probability that this happens AND I get an orange and Annie gets an apple is 15/91. 

 

Case 1.3, an apple and two oranges. The number of ways this can happen is 10 times 5 choose 2 = 100. So the probability that this happens 100/455 = 20/91. The probability that I get an orange is 2/3 and the probability that she gets an apple after that is 1/2, multiplying to 1/3. So the probability that both Case 1.3 occurs AND I get an orange and Annie gets an apple is 20/273. 

 

Case 1.4, three oranges. The number of ways to do this is 5 choose 3 = 10. So the probability of this occurring is 2/91. The probability that I get an orange and she gets an apple is 0, since there are no apples. 

 

Now let's move on to Case 2, 9 apples and 6 oranges. 

 

Case 2.1, three apples. The number of ways that this can happen is 9 choose 3, which is equal to 84. So the probability for this occurring is 84/455 = 12/65. After this, the probability that I get an orange and she gets an apple is 0, since there are no oranges. 

 

Case 2.2, two apples and an orange. The number of ways that this can happen is 9 choose 2 times 6 = 216, since we're choosing 2 apples and an orange. So the probability that this will happen is 216/455. The probability that I get an orange is 1/3 and the probability that she gets an apple after that is 1. Then the probability that this happens AND I get an orange and Annie gets an apple is 72/455. 

 

Case 2.3, an apple and two oranges. The number of ways this can happen is 9 times 6 choose 2 = 135. So the probability that this happens is 135/455 = 27/91. The probability that I get an orange is 2/3 and the probability that she gets an apple after that is 1/2, multiplying to 1/3. So the probability that both Case 1.3 occurs AND I get an orange and Annie gets an apple is 9/91.  

 

Case 2.4, three oranges. The number of ways to do this is 6 choose 3 = 20. So the probability of this occurring is 4/91. The probability that I get an orange and she gets an apple is 0, since there are no apples. 

 

So now we add all the probabilities from Case 1 and multiply by 5/8 since the probability of Case 1 occurring is 5/8, getting 5/8(20/273 + 15/91) = 25/168, and with Case 2 we get 3/8(72/455 + 9/91) = 27/280. Adding them, we obtain 103/420. 

I am not entirely sure this answer is correct, please correct me if it's wrong. 

 

Actually the answer cannot be determined

You have not specified where you got the money from to buy the fruits. Also, no info is given on allergies/ preferences, so for all we know, Annie could be allergic to oranges or you may hate apples

Annie's screwed then, I guess. :P

Edited December 28, 2016 by basenji7

[yonatanthesuper 314]

 

Ok, so I have 5 apples and 3 oranges. There is a 50% chance that I ate an apple and a 50% change that I ate an orange. Then I bought 5 apples and 3 oranges again. At this point, the probability of having 10 apples and 5 oranges is 1/2 and the probability of having 9 apples and 6 oranges is 1/2. 

 

 

Let's start with Case 1, 10 apples and 5 oranges. There are four possibilities of the fruits I gave Annie, three apples, two, one, or none. The total amount of ways to choose the fruits I give Annie is 15 choose 3, which is 455. 

 

Case 1.1, three apples. The number of ways that this can happen is 10 choose 3, which is equal to 120. So the probability for this occurring 120/455 = 24/91. After this, the probability that I get an orange and she gets an apple is 0, since there are no oranges. 

 

Case 1.2, two apples and an orange. The number of ways that this can happen is 10 choose 2 times 5 = 225, since we're choosing 2 apples and an orange. So the probability that this will happen is 45/91. The probability that I get an orange is 1/3 and the probability that she gets an apple after that is 1. Then the probability that this happens AND I get an orange and Annie gets an apple is 15/91. 

 

Case 1.3, an apple and two oranges. The number of ways this can happen is 10 times 5 choose 2 = 100. So the probability that this happens 100/455 = 20/91. The probability that I get an orange is 2/3 and the probability that she gets an apple after that is 1/2, multiplying to 1/3. So the probability that both Case 1.3 occurs AND I get an orange and Annie gets an apple is 20/273. 

 

Case 1.4, three oranges. The number of ways to do this is 5 choose 3 = 10. So the probability of this occurring is 2/91. The probability that I get an orange and she gets an apple is 0, since there are no apples. 

 

Now let's move on to Case 2, 9 apples and 6 oranges. 

 

Case 2.1, three apples. The number of ways that this can happen is 9 choose 3, which is equal to 84. So the probability for this occurring is 84/455 = 12/65. After this, the probability that I get an orange and she gets an apple is 0, since there are no oranges. 

 

Case 2.2, two apples and an orange. The number of ways that this can happen is 9 choose 2 times 6 = 216, since we're choosing 2 apples and an orange. So the probability that this will happen is 216/455. The probability that I get an orange is 1/3 and the probability that she gets an apple after that is 1. Then the probability that this happens AND I get an orange and Annie gets an apple is 72/455. 

 

Case 2.3, an apple and two oranges. The number of ways this can happen is 9 times 6 choose 2 = 135. So the probability that this happens is 135/455 = 27/91. The probability that I get an orange is 2/3 and the probability that she gets an apple after that is 1/2, multiplying to 1/3. So the probability that both Case 1.3 occurs AND I get an orange and Annie gets an apple is 9/91.  

 

Case 2.4, three oranges. The number of ways to do this is 6 choose 3 = 20. So the probability of this occurring is 4/91. The probability that I get an orange and she gets an apple is 0, since there are no apples. 

 

So now we add all the probabilities from Case 1 and divide them by 2 since the probability of Case 1 occurring is 1/2, getting 1/2(20/273 + 15/91) = 5/42, and with Case 2 we get 1/2(72/455 + 9/91) = 9/70. Adding them, we obtain 26/105. 

I am not entirely sure this answer is correct, please correct me if it's wrong. 

 

Annie's screwed then, I guess. :P

 

:blink: bro,you gotta get a life no offence.

[delnina2 1,570]

 

Ok, so I have 5 apples and 3 oranges. There is a 50% chance that I ate an apple and a 50% change that I ate an orange. Then I bought 5 apples and 3 oranges again. At this point, the probability of having 10 apples and 5 oranges is 1/2 and the probability of having 9 apples and 6 oranges is 1/2. 

 

 

Let's start with Case 1, 10 apples and 5 oranges. There are four possibilities of the fruits I gave Annie, three apples, two, one, or none. The total amount of ways to choose the fruits I give Annie is 15 choose 3, which is 455. 

 

Case 1.1, three apples. The number of ways that this can happen is 10 choose 3, which is equal to 120. So the probability for this occurring 120/455 = 24/91. After this, the probability that I get an orange and she gets an apple is 0, since there are no oranges. 

 

Case 1.2, two apples and an orange. The number of ways that this can happen is 10 choose 2 times 5 = 225, since we're choosing 2 apples and an orange. So the probability that this will happen is 45/91. The probability that I get an orange is 1/3 and the probability that she gets an apple after that is 1. Then the probability that this happens AND I get an orange and Annie gets an apple is 15/91. 

 

Case 1.3, an apple and two oranges. The number of ways this can happen is 10 times 5 choose 2 = 100. So the probability that this happens 100/455 = 20/91. The probability that I get an orange is 2/3 and the probability that she gets an apple after that is 1/2, multiplying to 1/3. So the probability that both Case 1.3 occurs AND I get an orange and Annie gets an apple is 20/273. 

 

Case 1.4, three oranges. The number of ways to do this is 5 choose 3 = 10. So the probability of this occurring is 2/91. The probability that I get an orange and she gets an apple is 0, since there are no apples. 

 

Now let's move on to Case 2, 9 apples and 6 oranges. 

 

Case 2.1, three apples. The number of ways that this can happen is 9 choose 3, which is equal to 84. So the probability for this occurring is 84/455 = 12/65. After this, the probability that I get an orange and she gets an apple is 0, since there are no oranges. 

 

Case 2.2, two apples and an orange. The number of ways that this can happen is 9 choose 2 times 6 = 216, since we're choosing 2 apples and an orange. So the probability that this will happen is 216/455. The probability that I get an orange is 1/3 and the probability that she gets an apple after that is 1. Then the probability that this happens AND I get an orange and Annie gets an apple is 72/455. 

 

Case 2.3, an apple and two oranges. The number of ways this can happen is 9 times 6 choose 2 = 135. So the probability that this happens is 135/455 = 27/91. The probability that I get an orange is 2/3 and the probability that she gets an apple after that is 1/2, multiplying to 1/3. So the probability that both Case 1.3 occurs AND I get an orange and Annie gets an apple is 9/91.  

 

Case 2.4, three oranges. The number of ways to do this is 6 choose 3 = 20. So the probability of this occurring is 4/91. The probability that I get an orange and she gets an apple is 0, since there are no apples. 

 

So now we add all the probabilities from Case 1 and divide them by 2 since the probability of Case 1 occurring is 1/2, getting 1/2(20/273 + 15/91) = 5/42, and with Case 2 we get 1/2(72/455 + 9/91) = 9/70. Adding them, we obtain 26/105. 

I am not entirely sure this answer is correct, please correct me if it's wrong. 

 

Annie's screwed then, I guess. :P

 

You have a little mistake in the first sentence.  If I have 5 apples and 3 ornages  a 62.5% chance that I ate an apple and a 37.5% change that I ate an orange. (I have 8 fruits in total five are apples and three of them are ornages. 5/8 for the apples. and 3/8 for the ornages. 

[Tofu 6,773]

:blink: bro,you gotta get a life no offence.

She's a girl

[personia 4,036]

:blink: bro,you gotta get a life no offence.

Oi dude, can you please tell us what the answer is to number 1? That seems to be the question that nobody can answer.

[yonatanthesuper 314]

Oi dude, can you please tell us what the answer is to number 1? That seems to be the question that nobody can answer.

sorry, i don't know,even my genius teacher didn't know.i will ask my math teacher after hanuka break.

Edited December 26, 2016 by yonatanthesuper

[personia 4,036]

sorry, i don't know,even my genius teacher didn't know.i will ask my math teach after hanuka break.

Okay, thank you.

[BlueDragon_tamki 3,204]

Damn too much math here

[gudia.zn 38]

i know answer no. 3 both r pounds so both r of same weight :D

Edited December 27, 2016 by gudia.zn

[Tofu 6,773]

2 pounds of feathers is heavier because you have to carry the weight of what you did to those poor birds as well.

 

 

i know answer no. 3 both r pounds so both r of same weight :D

[basenji7 1,199]

You have a little mistake in the first sentence.  If I have 5 apples and 3 ornages  a 62.5% chance that I ate an apple and a 37.5% change that I ate an orange. (I have 8 fruits in total five are apples and three of them are ornages. 5/8 for the apples. and 3/8 for the ornages. 

Ahh I missed that :x 

not hard to fix though 

 

She's a girl

I am. :P

 

And I sincerely apologize for my lack of a life, but unfortunately I am never going to get a life. Also I cannot help it when I see a math problem I can't not do it. 

Edited December 28, 2016 by basenji7

[BlueDragon_tamki 3,204]

Ahh I missed that :x

not hard to fix though

 

I am. :P

 

And I sincerely apologize for my lack of a life, but unfortunately I am never going to get a life. Also I cannot help it when I see a math problem I can't not do it.

most of my classmates are like this too