Probability

[BrownieDDD]

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Found this in my math destroyer questions:

A bag contains 7 walnuts, 4 cashews, and 3 pitachio nuts. 2 of the nuts fall out at random. Wah tis the probability that one is a cashew and the other is a walnut?

answer: (7/14) (4/13) + (4/14) (7/13)

So you pick one out, and don't replace it...that's what I got from reading the solution. However, when i first read it, i thought they fall out at the same time (kind of similar to picking out 2 at the same time). so I did (7/14) (4/14)



2nd probability question

What is the probability of rolling a sum of six and then a sum of six again in 2 consectuive rolls of a pair of dice?

answer: 5^2/6^4

 

[sacapuntas]

2nd probability question

What is the probability of rolling a sum of six and then a sum of six again in 2 consectuive rolls of a pair of dice?

answer: 5^2/6^4

I think the easiest way to think about this one is this:

There are 5 possibilities of rolling 6 -> (1&5, 2&4, 3&3, 4&2, 5&1) and 36 total possibilities. That means on the first roll your odds are 5/36. Your odds on the second roll are also 5/36. Multiply these 2 together, ((5/36)*(5/36) and it is actually the same thing as (5*5)/(6*6*6*6) which is the same thing as your given answer.

 

[BrownieDDD]

thanks, much appreciated. As for the 2nd one, was I completely wrong in thinking both can fall out at the same time? So that the chances of picking a walnut and cashew are as foolllows 7/14 (4/14)

 

[ushaseos]

The "without replacement" is the clue that they don't fall out at the same time. Everytime you see without replacement, your denominators should start to look like a partial factorial (14,13,12...). In this particular case since order isn't specific it could be either cashew first and walnut second or vice-versa so you add the two probabilities together.

 

[Jay20155555]

I think the easiest way to think about this one is this:

There are 5 possibilities of rolling 6 -> (1&5, 2&4, 3&3, 4&2, 5&1) and 36 total possibilities. That means on the first roll your odds are 5/36. Your odds on the second roll are also 5/36. Multiply these 2 together, ((5/36)*(5/36) and it is actually the same thing as (5*5)/(6*6*6*6) which is the same thing as your given answer.

Why wouldnt there be 6 possibilities for 3,3 and 3,3? Each 3 from a different die

 

[adria_justine]

I think the easiest way to think about this one is this:

There are 5 possibilities of rolling 6 -> (1&5, 2&4, 3&3, 4&2, 5&1) and 36 total possibilities. That means on the first roll your odds are 5/36. Your odds on the second roll are also 5/36. Multiply these 2 together, ((5/36)*(5/36) and it is actually the same thing as (5*5)/(6*6*6*6) which is the same thing as your given answer.