Odds of Getting 2 Out of 3 Correct

[swerdna]

With three 50/50 odds choices, what are the odds of getting any 2 correct?

 

[berkeman]

With three 50/50 odds choices, what are the odds of getting any 2 correct?

What are your thoughts? It also depends on whether each choice is independent of the previous choices (like coin flips are). This is a bit too much like homework/coursework, so I'm moving it to Homework Help.

 

[swerdna]

What are your thoughts? It also depends on whether each choice is independent of the previous choices (like coin flips are). This is a bit too much like homework/coursework, so I'm moving it to Homework Help.

I'm far too old for it to be homework. Each choice is independant. If a person was guessing the colour of randomly presented unseen playing cards, what are the chaces of guessing any two of the thee correctly. I think it must be less than 1 in 4.

 

[berkeman]

I'm far too old for it to be homework. Each choice is independant. If a person was guessing the colour of randomly presented unseen playing cards, what are the chaces of guessing any two of the thee correctly. I think it must be less than 1 in 4.

Hey, I'm pretty old as well, and do lots of homework!

The key is to think of it as a decision tree. You can write it out like this, with a correct pick = 1, and a wrong pick = 0:

1st 2nd 3rd Total
 0    0    0     0
 0    0    1     1
 0    1    0     1
 0    1    1     2
 1    0    0     1
 1    0    1     2
 1    1    0     2
 1    1    1     3

There are 8 possible outcomes. How many of them result in getting 2 right?

Can you see how you would extend this to, say, a 60/40 chance of picking correctly?

 

[swerdna]

Hey, I'm pretty old as well, and do lots of homework!

The key is to think of it as a decision tree. You can write it out like this, with a correct pick = 1, and a wrong pick = 0:

1st 2nd 3rd Total
 0    0    0     0
 0    0    1     1
 0    1    0     1
 0    1    1     2
 1    0    0     1
 1    0    1     2
 1    1    0     2
 1    1    1     3

There are 8 possible outcomes. How many of them result in getting 2 right?

Can you see how you would extend this to, say, a 60/40 chance of picking correctly?

I have it - Thanks

 

[BoundByAxioms]

If you've learned the chose formula, then a good way to do problems that have specifically 50/50 odds is this:
\frac{n C r}{2^n} where n is the number of trials, and r is the number of successes. So, to answer your question, 3 C 2=3, and 2³=8. So your answer should be \frac{3}{8}