If 3 coins are thrown, the probability to have exactly two heads is:
a) 3/8
b) 2/6
Explain.
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?.
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.