Probability Question: Cubing 1/3 for 3 Die?

[DarkStalker]

1. Question http://tinyurl.com/ydwpqx4

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Hello. The link I provided has the entire statement of the question, and it's solution (Example 4.8).
We know that P(1 or 6) for one throw = 1/3, and P(neither 1 nor 6) = 2/3.
What I don't understand is why we don't cube 1/3 to get the probability of getting 1 or 6 for 3 die. The solution first cubes 2/3 and then subtracts it from the total probability 1. If we can get the probability of not getting 1 or 6 by cubing it's probability, why can't we get the the probability of getting 1 or 6 by cubing it's probability?

Thanks.

 

[CompuChip]

(1/3)³ is the probability that you get 1 or 6, on all three throws.

What you want is the probability that you get 1 or 6 on at least one of them. So you could write out the list (where Y means throwing 1 or 6, and N means any other number):
P(1 or 6 shows up) = P(first throw gives one or 6) + P(second throw gives 1 or 6) + P(third throw gives 1 or 6) = P(Y) + P(NY) + P(NNY) = (1/3) + (2/3)(1/3) + (2/3)(2/3)(1/3) = (1/3)(1 + 2/3 + 4/9) = (1/3)(19/9) = 19/27.

The solutions take a shortcut, using the complement rule.

 

[DarkStalker]

(1/3)³ is the probability that you get 1 or 6, on all three throws.

What you want is the probability that you get 1 or 6 on at least one of them.

Thanks a lot for your reply, and you cleared up the problem for me. But...

So you could write out the list (where Y means throwing 1 or 6, and N means any other number):
P(1 or 6 shows up) = P(first throw gives one or 6) + P(second throw gives 1 or 6) + P(third throw gives 1 or 6) = P(Y) + P(NY) + P(NNY) = (1/3) + (2/3)(1/3) + (2/3)(2/3)(1/3) = (1/3)(1 + 2/3 + 4/9) = (1/3)(19/9) = 19/27.

The solutions take a shortcut, using the complement rule

This is where I lose you. Wouldn't the probability of getting 1 or 6 in the second throw be 1/3 as well?

 

[CompuChip]

Yes.

But what I mean by P(second throw gives 1 or 6) is actually: P(second throw gives 1 or 6 and first throw doesn't).
Since the throws are independent of one another, this is

P(second throw gives 1 or 6) x P(first throw doesn't give 1 or 6) = (1/3) x (2/3).

This counts the possibilities YY, YN (from the first 1/3) and NY (from the 1/3 x 2/3).

Just writing 1/3 also includes the possibility of both the first and the second giving 1 or 6, and you would count: YY, YN (from the first 1/3) and NY, YY (from the second 1/3). You see that you are double-counting YY (1 or 6 in both throws).

 

[DarkStalker]

Yes.

But what I mean by P(second throw gives 1 or 6) is actually: P(second throw gives 1 or 6 and first throw doesn't).
Since the throws are independent of one another, this is

P(second throw gives 1 or 6) x P(first throw doesn't give 1 or 6) = (1/3) x (2/3).

This counts the possibilities YY, YN (from the first 1/3) and NY (from the 1/3 x 2/3).

Just writing 1/3 also includes the possibility of both the first and the second giving 1 or 6, and you would count: YY, YN (from the first 1/3) and NY, YY (from the second 1/3). You see that you are double-counting YY (1 or 6 in both throws).

I get it now. Thanks a lot, you're a life saver!