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 :)

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.

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

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,

Re: Coin toss probability question

The joint probability of no heads is 1/8.

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?

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?

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?**

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

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?

Re: Coin toss probability question

Re: Coin toss probability question

Quote:

Originally Posted by

**Schuyla** 2 have exactly 1 head

**But 3 have exactly the**__ same number__ of heads.

Re: Coin toss probability question

Okay, I understand now. I can see where there are the three are. So the answer to the question would be 3/8 correct?