# Thread: Irritating probability question

1. ## Irritating probability question

A Man has three coins one of which is biased so that it turns up tails 80% of the time.If he randomly selects one of the coins, tosses it three times, and obtains three tails, what is the probability that this is the biased coin?

By Using Bayes Theorem the answer is .672. What are the steps to getting this answer???

2. let $\displaystyle T$ denote the event that 3 tails are obtained after 3 tosses of the coin

and let $\displaystyle B$ denote the event that the coin is biased

so the event that the coin is not biased is $\displaystyle B^C$

now, the question is telling you to find that you find if the coin is biased given that 3 tails occur

So, using Bayes theorem you can write:

$\displaystyle P(B|T) = \frac{P(T|B) \times P(B)}{P(T|B)\times P(B) + P(T|B^{C})\times P(B^{C}) }$

Try plugging in the appropriate numbers here!!

3. Hello, Aquameatwad!

A man has three coins one of which is biased so that it turns up tails 80% of the time.
If he randomly selects one of the coins, tosses it three times, and obtains three tails,
what is the probability that this is the biased coin?

We want: .$\displaystyle P(\text{biased}\,|\,\text{3 Tails})$

Bayes' Theorem: .$\displaystyle P(\text{biased}\,|\,\text{3 Tails}) \;=\;\dfrac{P(\text{biased}\,\wedge\,\text{3 Tails})}{P(\text{3 Tails})}$ [1]

The numerator is: .$\displaystyle P(\text{biased}\,\wedge\,\text{3 Tails})$

. . $\displaystyle \begin{array}{ccc}P(\text{biased}) &=& \frac{1}{3} \\ P(\text{3 Tails}) &=& \left(\frac{4}{5}\right)^3 \end{array}$

The numerator is: .$\displaystyle \dfrac{1}{3}\left(\dfrac{4}{5}\right)^3 \:=\:\dfrac{64}{375}$

The denominator is: .

$\displaystyle P(\text{3 Tails}) \;=\;P(\text{unbiased}\,\wedge\,\text{3 Tails}) + P(\text{biased}\,\wedge\,\text{3 Tails})$

. . $\displaystyle \begin{array}{ccc}P(\text{unbiased}) &=& \frac{2}{3} \\ P(\text{3 Tails}) &=& \left(\frac{1}{2}\right)^3 \end{array}$
. . $\displaystyle P(\text{unbiased}\,|\,\text{3 Tails}) \;=\;\frac{2}{3}\left(\frac{1}{2}\right)^3 \;=\;\frac{1}{12}$

The denominator is: .$\displaystyle \dfrac{1}{12} + \dfrac{64}{375} \;=\;\dfrac{127}{500}$

Then [1] becomes:

. . $\displaystyle P(\text{biased}\,|\text{3 tails}) \;=\;\dfrac{\frac{64}{375}}{\frac{127}{500}} \;=\;\dfrac{256}{381} \;=\;0.67191601 \;\approx\;0.672$