Conditional Probability Question

[dxoo]

A coin is tossed three times, let E be the event we get at most two tails, let F be the event we get at least one tail. Find P(E | F)

My Approach:
[MATH] P(E) = P(1 tail) + P(2 tails)\\ P(E) = {\dfrac{({3 \choose 1} + {3 \choose 2})}{2^3}}\\ P(E) = {\dfrac{6}{8}}\\ [/MATH][MATH] \\P(F) = 1 - P(HHH)\\ P(F) = 1- 1/8\\ P(F) = {\dfrac{7}{8}}\\ [/MATH]
Now, [MATH] P(E|F) = {\dfrac{E \cap F}{F}}\\ P({E \cap F}) = 6/8 + 7/8 - 1\\ P({E \cap F}) = {\dfrac{5}{8}} \\ \\ \\ P({\dfrac{E \cap F}{F}}) = {\dfrac{5/8}{7/8}}\\ P({\dfrac{E \cap F}{F}}) = {\dfrac{5}{7}}\\ [/MATH]

However, the correct answer is [MATH]{\dfrac{6}{7}}\\[/MATH]. So I know I've made an error in calculating the intersection of E and F, but why is my technique wrong, and what would be the correct way?

Any help is appreciated

 

[Romsek]

if [MATH]E[/MATH] is the event at most 2 tails it should include the 0 tails case as well.

it's not until you incorporate that [MATH]F[/MATH] is given, i.e. [MATH]P[E|F][/MATH] that you can remove that case.

It's probably easiest just to calculate [MATH]P[E|F] = \dfrac{P[E \cap F]}{P[F]}[/MATH] directly.

[MATH]\dfrac{P[E \cap F]}{P[F]} = \dfrac{P[1]+P[2]}{1-P[0]} = \dfrac{\frac 3 8 + \frac 3 8}{1 - \frac 1 8} = \dfrac 6 7 [/MATH]

 

[dxoo]

if [MATH]E[/MATH] is the event at most 2 tails it should include the 0 tails case as well.

it's not until you incorporate that [MATH]F[/MATH] is given, i.e. [MATH]P[E|F][/MATH] that you can remove that case.

It's probably easiest just to calculate [MATH]P[E|F] = \dfrac{P[E \cap F]}{P[F]}[/MATH] directly.

[MATH]\dfrac{P[E \cap F]}{P[F]} = \dfrac{P[1]+P[2]}{1-P[0]} = \dfrac{\frac 3 8 + \frac 3 8}{1 - \frac 1 8} = \dfrac 6 7 [/MATH]

That makes a lot of sense, thank you!

 

[Steven G]

HHH
THH
HTH
HHT
HTT
THT
TTH
TTT x
Is the answer clear now? This method always works. Then you can try to use a formula to make sure that you get the correct formula.