1. 3 coins

If 3 coins are thrown, the probability to have exactly two heads is:

a) 3/8
b) 2/6

Explain.

2. $\displaystyle \begin{pmatrix}3\\2\end{pmatrix}p^{2}q$

Where p=q=.5

You could try expanding $\displaystyle (p+q)^{3}$ and looking at the binomial expansion.

3. Is there an easier way to find out the answer, is it that hard?

4. That isn't difficult.

It may look imposing if you don't understand the notation.

The number of ways to choose 2 out of 3 objects is 3.

You have $\displaystyle 3p^{2}q$

Enter in your probabilities of success and failure. If you flip a coin the probabilities are 50-50.

So, you have $\displaystyle 3(\frac{1}{2})^2(\frac{1}{2})=\frac{3}{8}$

Maybe I should've explained it this way to start with:

Throw the 3 coins. How many possible arrangements can you have?.

$\displaystyle 2^{3}=8$. Right?.

Let's list them:

HTT
HHH
TTT
HHT
THH
HTH
THT
TTH

Now, how many of those have exactly 2 H's?.....3. See?.

5. Hello, Bret!

Here's yet another approach . . .

If 3 coins are thrown, the probability to have exactly two heads is:

. . . $\displaystyle a)\;\frac{3}{8}\qquad b)\:\frac{2}{6}$

Explain.

Suppose we want the coins to turn up $\displaystyle H\!HT$ in that order.

Prob. that coin #1 is Heads: $\displaystyle \frac{1}{2}$
Prob. that coin #2 is Heads: $\displaystyle \frac{1}{2}$
Prob. that coin #3 is Tails: $\displaystyle \frac{1}{2}$

Hence: .$\displaystyle P(HHT) \: =\:\left(\frac{1}{2}\right)^3\:=\:\frac{1}{8}$

Since the problem did not insist on a specific order of Heads and Tails,
. . there are three ways to get two Heads: .$\displaystyle HHT,\;HTH,\;THH$

Therefore: .$\displaystyle P(\text{2H, 1T})\;=\;3 \times \frac{1}{8}\;= \; \boxed{\frac{3}{8}}$

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A handy tip

If there are $\displaystyle 3$ coins, how can we list all the outcomes?

Since each coin can turn up in 2 ways, there are: $\displaystyle 2^3 = 8$ outcomes.

Write $\displaystyle 8\;H's\text{ and }T's$ . . . first half $\displaystyle H$'s, the second half $\displaystyle T$is.

. . . $\displaystyle \left \begin{array}{cccc}H\\H\\H\\H\end{array}\;\right\}$
. . . $\displaystyle \left \begin{array}{cccc}T\\T\\T\\T\end{array}\;\right\}$

Write $\displaystyle 8\;H's\text{ and }T's$ . . . this time "change every two".

. . . $\displaystyle \left \begin{array}{cc}H\\H\end{array}\;\right\}$
. . . $\displaystyle \left \begin{array}{cc}T\\T\end{array}\;\right\}$
. . . $\displaystyle \left \begin{array}{cc}H\\H\end{array}\;\right\}$
. . . $\displaystyle \left \begin{array}{cc}T\\T\end{array}\;\right\}$

Write $\displaystyle 8\;H's\text{ and }T's$ . . . this time "change every one" (alternate).

. . . $\displaystyle \begin{array}{cccc}\left H\;\;\right\} \\ \left T\;\;\right\} \\ \left H\;\;\right\} \\ \left T\;\;\right\}\end{array}$
. . . $\displaystyle \begin{array}{cccc} \left H\;\;\right\} \\ \left T\;\;\right\} \\ \left H\;\;\right\} \\ \left T\;\;\right\}\end{array}$

Write the three columns side-by-side:

. . . $\displaystyle \begin{array}{cccccccc}1)&H&H&H\\2)&H&H&T\\3)&H&T& H\\4)&H&T&T\\5)&T&H&H\\6)&T&H&T\\7)&T&T&H\\8)&T&T& T\end{array}$

. . . and we have the eight possible outcomes!

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Of course, this procedure works for 4 coins (16 outcomes)
. . and 5 coins (32 outcomes) . . . it just takes longer.

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