# Fair and 2 headed coins

• Oct 22nd 2009, 12:27 PM
bjanela

You have 2 fair coins and one two headed coin. You draw one coin randomly and flip and toss it twice. Given that both tosses resulted in heads, find the conditional probability that the two headed coin was chosen as a fraction--in lowest terms.

We also have to do this again, but assume thaqt we have 8 fair coins and one 2-headed coin.

Thanks!
• Oct 22nd 2009, 02:45 PM
tonio
Quote:

Originally Posted by bjanela

You have 2 fair coins and one two headed coin. You draw one coin randomly and flip and toss it twice. Given that both tosses resulted in heads, find the conditional probability that the two headed coin was chosen as a fraction--in lowest terms.

We also have to do this again, but assume thaqt we have 8 fair coins and one 2-headed coin.

Thanks!

Put $\displaystyle A=\mbox{ the two-headed coin was chosen, and }$ $\displaystyle B=\mbox{ we get two heads after flipping a coin twice}$

The conditional probability we're looking for is:

$\displaystyle P\left(A\backslash B\right)=\frac{P\left(A \cap B\right)}{P(B)}$

But $\displaystyle P\left(A \cap B\right)=\frac{1}{3}$, so you only need the denominator above which is pretty easy (construct a two-stage 3-branched probability tree)

Tonio
• Oct 22nd 2009, 11:13 PM
bjanela
Would P(B) = 1/4? The result of the prob. tree being (HH out of HH, HT, TH, TT)? If so, then P(A|B) = 4/3 which is impossible =(
• Oct 23rd 2009, 10:46 AM
Soroban
Hello, bjanela!

We need Bayes' Theorem: .$\displaystyle P(A\,|\,B) \:=\:\frac{P(A\wedge B)}{P(B)}$

Quote:

You have 2 fair coins $\displaystyle (f)$ and one two-headed coin $\displaystyle (d)$.
You draw one coin randomly and flip and toss it twice.
Given that both tosses resulted in heads, find the conditional probability
that the two-headed coin was chosen.

We want: .$\displaystyle P(d\,|\,HH) \;=\;\frac{P(d\wedge HH)}{P(HH)}$

A fair coin is chosen: .$\displaystyle P(f) \:=\:\tfrac{2}{3}$
. . Then: .$\displaystyle P(HH) \:=\:\tfrac{1}{4}$
. . Hence: .$\displaystyle P(f \wedge HH) \:=\:\tfrac{2}{3}\cdot\tfrac{1}{4} \:=\:\tfrac{1}{6}$

$\displaystyle d$ is chosen: .$\displaystyle P(d) \:=\:\tfrac{1}{3}$
. . Then: .$\displaystyle P(HH) \:=\:1$
. . Hence: .$\displaystyle P(d \wedge HH) \:=\:\tfrac{1}{3}$

So: .$\displaystyle P(HH) \:=\:\tfrac{1}{6} + \tfrac{1}{3} \:=\:\tfrac{1}{2}$

Therefore: .$\displaystyle P(d\,|\,HH) \:=\:\frac{P(d\,\wedge\,HH)}{P(HH)} \:=\:\frac{\frac{1}{3}}{\frac{1}{2}} \:=\:\frac{2}{3}$

Quote:

Do this again, with 8 fair coins and one 2-headed coin.

We want: .$\displaystyle P(d\,|\,HH) \;=\;\frac{P(d\wedge HH)}{P(HH)}$

A fair coin is chosen: .$\displaystyle P(f) \:=\:\tfrac{8}{9}$
. . Then: .$\displaystyle P(HH) \:=\:\tfrac{1}{4}$
. . Hence: .$\displaystyle P(f \wedge HH) \:=\:\tfrac{8}{9}\cdot\tfrac{1}{4} \:=\:\tfrac{2}{9}$

$\displaystyle d$ is chosen: .$\displaystyle P(d) \:=\:\tfrac{1}{9}$
. . Then: .$\displaystyle P(HH) \:=\:1$
. . Hence: .$\displaystyle P(d \wedge HH) \:=\:\tfrac{1}{9}$

So: .$\displaystyle P(HH) \:=\:\tfrac{2}{9} + \tfrac{1}{9} \:=\:\tfrac{1}{3}$

Therefore: .$\displaystyle P(d\,|\,HH) \:=\:\frac{P(d\,\wedge\,HH)}{P(HH)} \:=\:\frac{\frac{1}{9}}{\frac{1}{3}} \:=\:\frac{1}{3}$