# Coin toss probability question

• Apr 4th 2012, 11:12 AM
Schuyla
Coin toss probability question
Al tosses one quarter and at the same time Betty tosses two quarters. What is the probability that Betty gets the same number of heads as Al?

This is what I have:
Al H/T ½ 50%
Betty HH / HT / TH/ TT ½ 50%
The probability is ½ or 50% chance that Betty will get the same amount of heads as Al

Can you see if I'm on the right track or can you guide me to the right track :)
• Apr 4th 2012, 11:50 AM
Plato
Re: Coin toss probability question
Quote:

Originally Posted by Schuyla
Al tosses one quarter and at the same time Betty tosses two quarters. What is the probability that Betty gets the same number of heads as Al?

Well, these are independent.
They are only two values: both get zero heads or both get exactly one head.
Find the probability of both of those for each, multiply and add together.
• Apr 4th 2012, 05:00 PM
Schuyla
Re: Coin toss probability question
If I am understanding right both of the probabilities
1/2 * 1/2 = 1/4?
Since there is a 50/50% chance Al will get a head, and there is also a 50/50% chance Betty will get heads on at least on coin. So what I do is what is above, correct?
Schuyla
• Apr 5th 2012, 04:08 AM
Plato
Re: Coin toss probability question
Quote:

Originally Posted by Schuyla
If I am understanding right both of the probabilities 1/2 * 1/2 = 1/4? Since there is a 50/50% chance Al will get a head, and there is also a 50/50% chance Betty will get heads on at least on coin. So what I do is what is above, correct?

Have you read the question? It says that they get the same number of heads.
So that is either zero or one.
The probability he gets no head is $\displaystyle \frac{1}{2}$. The probability she gets no head is $\displaystyle \frac{1}{4}$.
So what is the joint probability of no heads?

Now the same for exactly one head each,
• Apr 6th 2012, 07:37 AM
Schuyla
Re: Coin toss probability question
The joint probability of no heads is 1/8.
• Apr 6th 2012, 07:49 AM
Plato
Re: Coin toss probability question
Quote:

Originally Posted by Schuyla
The joint probability of no heads is 1/8.

Correct!
Now what is the probability of exactly one head for each?
• Apr 6th 2012, 07:53 AM
Schuyla
Re: Coin toss probability question
Will if the probability of Al getting heads is 1/2 and the probability for Betty getting one head is 1/4. The probability of both of them getting 1 head is 1/8.
Okay I'm second guessing myself but with the previous answer I gave you of no heads being 1/8, would that make the possibility 7/8 for a head?
• Apr 6th 2012, 08:02 AM
Plato
Re: Coin toss probability question
Quote:

Originally Posted by Schuyla
Will if the probability of Al getting heads is 1/2 and the probability for Betty getting one head is 1/4. The probability of both of them getting 1 head is 1/8.
Okay I'm second guessing myself but with the previous answer I gave you of no heads being 1/8, would that make the possibility 7/8 for a head?

NO! The probability for Betty getting exactly one head is $\displaystyle \frac{1}{2}$. WHY?
• Apr 6th 2012, 08:06 AM
Schuyla
Re: Coin toss probability question
If the probability for Betty getting exactly one head is 1/2 is because she has two quarters and there is a 50% chance that she would get at least one head.

Betty possible outcomes are {HH, HT, TH, TT} 2 out of the 4 tosses have 1 head which equals 1/2
• Apr 6th 2012, 08:17 AM
Plato
Re: Coin toss probability question
Quote:

Originally Posted by Schuyla
If the probability for Betty getting exactly one head is 1/2 is because she has two quarters and there is a 50% chance that she would get at least one head.

Look, here is what that space looks like.
$\displaystyle \begin{array}{*{20}{c}} {Al}&\vline & {Betty} \\ \hline H&\vline & {HH} \\ H&\vline & {HT} \\ H&\vline & {TH} \\ H&\vline & {TT} \\ T&\vline & {HH} \\ T&\vline & {HT} \\ T&\vline & {TH} \\ T&\vline & {TT} \end{array}$
There are eight elementary events.
How many have the same number of heads?
• Apr 6th 2012, 08:20 AM
Schuyla
Re: Coin toss probability question