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

a) 3/8

b) 2/6

Explain.

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- Aug 4th 2006, 09:16 AMbret803 coins
If 3 coins are thrown, the probability to have exactly two heads is:

a) 3/8

b) 2/6

Explain. - Aug 4th 2006, 09:31 AMgalactus
$\displaystyle \begin{pmatrix}3\\2\end{pmatrix}p^{2}q$

Where p=q=.5

You could try expanding $\displaystyle (p+q)^{3}$ and looking at the binomial expansion. - Aug 5th 2006, 08:01 AMbret80
Is there an easier way to find out the answer, is it that hard?

- Aug 5th 2006, 08:13 AMgalactus
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?. - Aug 5th 2006, 09:21 AMSoroban
Hello, Bret!

Here's yet another approach . . .

Quote:

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!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Of course, this procedure works for 4 coins (16 outcomes)

. . and 5 coins (32 outcomes) . . . it just takes longer.