Conditional Probability...Special quarters?

[MathHelpPlease]

Hi Everybody!
I'm having some trouble doing this problem:
Mandvil has one standard quarter and one special quater with a Head on both sides. He selects one of these two coins at random, and without looking at it first, he flips the coin three times. If you flips a Head three straight times, what is the probability that he selected the special quarter? Express your answer as a common fraction.

I know that the answer is 8/9 because of "conditional probability" and how to arrive at the answer by using 1/[(8/9)+1] which makes 1/9/8=8/9
However, I'm just wondering why?
Could anyone help me?
Thanks a lot!

 

[mmm4444bot]

Hi MathHelpPlease:

1/(8/9 + 1) does not equal 1/9/8.

1/(8/9 + 1) = 9/17.

Regardless, will you please explain to me how you came up with the rational expression 1/(8/9 + 1).

Thanks.

~ Mark

 

[pka]

standard quarter and one special quater with a Head on both sides. He selects one of these two coins at random, and without looking at it first, he flips the coin three times. If you flips a Head three straight times, what is the probability that he selected the special quarter? Express your answer as a common fraction.
I know that the answer is 8/9

You have the correct answer, but the wrong reasoning and/or calculation.
If Q stands for the normal quarter, S for the special quarter and H for the three heads in a row, then you want to find \(\displaystyle P(S|H)\).
We know that \(\displaystyle P(H|S)=1\) because we can only have heads the special quarter.
But \(\displaystyle P(H|Q) = \frac{1}{8}\) as usual.
Because the coins a equally likely we can drop part of the usual formula.
Thus, \(\displaystyle P(S|H) = \frac{{P(S \cap H)}}{{P(H)}} = \frac{1}{{1 + \frac{1}{8}}} = \frac{8}{9}\).

 

[MathHelpPlease]

Oh, I'm sorry, I meant 1/8 instead of 8/9.
Thanks!