# Probabilities

• Feb 28th 2008, 07:12 AM
fulltwist8
Probabilities
Roll 2 dice and sum the spots. Then flip a coin that number of times. Find the exact probability of rolling a total of two if the experiment produced exactly as many heads as tails.

I recognize that it's a conditional probability but I don't know how to find the probability of getting exactly as many heads as tails.
• Feb 28th 2008, 07:30 AM
colby2152
Quote:

Originally Posted by fulltwist8
Roll 2 dice and sum the spots. Then flip a coin that number of times. Find the exact probability of rolling a total of two if the experiment produced exactly as many heads as tails.

I recognize that it's a conditional probability but I don't know how to find the probability of getting exactly as many heads as tails.

Probability of heads is the same as tails (50%).
• Feb 28th 2008, 07:57 AM
fulltwist8
but the probability of getting exactly the same number of heads and tails is not .5

That's what I'm having trouble finding
• Feb 28th 2008, 08:55 AM
Plato
Quote:

Originally Posted by fulltwist8
but the probability of getting exactly the same number of heads and tails is not .5

Exactly!
In order to have an equal number of head & tails, E, you must have rolled an even sum: 2 4 6 8 10 or 12.
Here is an example. The probability of E and a sum of 4 is
$\displaystyle P(E4) = P(E|4)P(4) = \left( {\frac{\binom {4}{2}}{{2^4 }}} \right)\left( {\frac{3}{{36}}} \right)$.

So you must work out the following:
$\displaystyle P(E) = P(E2) + P(E4) + P(E6) + P(E8) + P(E10) + P(E12)$.
That is your denominator in the answer to this question which is:
$\displaystyle P(2|E) = \frac{{P(E2)}}{{P(E)}}$ .
• Feb 28th 2008, 11:42 AM
Soroban
Hello, fulltwist8!

This will take a while . . .

Quote:

Roll two dice and sum the spots.
Then flip a coin that number of times.
Find the probability of rolling a total of 2
if there were exactly as many heads as tails.

We want:.$\displaystyle P[\text{(sum of 2) }|\text{ (H = T)}] \;=\; \frac{P[\text{(sum of 2)} \wedge \text{(H = T)}]} {P[\text{H = T}] }$

The numerator is: .$\displaystyle P[\text{(sum of 2) } \wedge \text{ (H = T)}]$

. . $\displaystyle P[\text{sum of 2}] \:=\:\frac{1}{36}$
. . With two tosses, $\displaystyle P[\text{H = T}] \:=\:\frac{2}{4} \:=\:\frac{1}{2}$
. . . . Hence: .$\displaystyle P[\text{(sum of 2) }\wedge\text{ (H = T)}] \;=\;\frac{1}{36}\cdot\frac{1}{2} \:=\:\frac{1}{72}$

The denominator is: .$\displaystyle P[\text{H = T}]$
We must consider all twelve oucomes of the dice.

$\displaystyle \begin{array}{ccccccccc} P[\text{sum of 2}] &=& \frac{1}{36} & & P[\text{H = T}] & = & \frac{1}{2} \\ P[\text{sum of 3}] &=& \frac{2}{36} & & P[\text{H = T}] &=& 0 \\ P[\text{sum of 4}] &=&\frac{3}{36} & & P[\text{H = T}] &=&{4\choose2}\left(\frac{1}{2}\right)^4 & = & \frac{3}{8} \\ P[\text{sum of 5}] &=&\frac{4}{36} & & P[\text{H = T}] &=& 0 \end{array}$
$\displaystyle \begin{array}{ccccccccc}P[\text{sum of 6}] &=&\frac{5}{36} & & P[\text{H = T}] &=&{6\choose3}\left(\frac{1}{2}\right)^6 &=& \frac{5}{16} \\ P[\text{sum of 7}] &=&\frac{6}{36} & & P[\text{H = T}] &=& 0 \\ P[\text{sum of 8}] &=&\frac{5}{36} & & P[\text{H = T}] &=& {8\choose4}\left(\frac{1}{2}\right)^8 &=& \frac{35}{128} \\ P[\text{sum of 9}] &=&\frac{4}{36} & & P[\text{H = T}] &=& 0 \end{array}$
$\displaystyle \begin{array}{ccccccccc}P[\text{sum of 10}] &=& \frac{3}{36} & & P[\text{H = T}] &=&{10\choose5}\left(\frac{1}{2}\right)^{10} &=& \frac{63}{256} \\ P[\text{sum of 11}] &=&\frac{2}{36} & & P[\text{H = T}] &=&0 \\ P[\text{sum of 12}] &=&\frac{1}{36} & & P[\text{H = T}] &=&{12\choose6}\left(\frac{1}{2}\right)^{12} &=& {\color{blue}\frac{231}{512}}\end{array}$

Hence: .$\displaystyle P[\text{H = T}] \;=\;\frac{1}{36}\!\cdot\!\frac{1}{2} + \frac{3}{36}\!\cdot\!\frac{3}{8} + \frac{5}{36}\!\cdot\!\frac{5}{16} + \frac{5}{36}\!\cdot\!\frac{35}{128} + \frac{3}{36}\!\cdot\!\frac{63}{256} + \frac{1}{36}\!\cdot\!\frac{231}{512} \;=\;\frac{2941}{18432}$

Therefore:.$\displaystyle P[\text{(sum of 2) }|\text{ (H = T)}] \;=\; \frac{P[\text{(sum of 2)} \wedge \text{(H = T)}]} {P[\text{H = T}] } \;=\;\frac{\frac{1}{72}}{\frac{2941}{18432}} \;=\;\boxed{\frac{256}{2941}}$

I hope my arithmetic is correct . . . this time
$\displaystyle \begin{array}{ccccccccc}P[\text{sum of 10}] &=& \frac{3}{36} & & P[\text{H = T}] &=&{10\choose5}\left(\frac{1}{2}\right)^{10} &=& \frac{63}{256} \\ P[\text{sum of 11}] &=&\frac{2}{36} & & P[\text{H = T}] &=&0 \\ P[\text{sum of 12}] &=&\frac{1}{36} & & P[\text{H = T}] &=&{12\choose6}\left(\frac{1}{2}\right)^{12} &=& \color{red}\frac{231}{1024}\end{array}$