Where p=q=.5
You could try expanding and looking at the binomial expansion.
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
Enter in your probabilities of success and failure. If you flip a coin the probabilities are 50-50.
So, you have
Maybe I should've explained it this way to start with:
Throw the 3 coins. How many possible arrangements can you have?.
. 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:
. . .
Explain.
Suppose we want the coins to turn up in that order.
Prob. that coin #1 is Heads:
Prob. that coin #2 is Heads:
Prob. that coin #3 is Tails:
Hence: .
Since the problem did not insist on a specific order of Heads and Tails,
. . there are three ways to get two Heads: .
Therefore: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
A handy tip
If there are coins, how can we list all the outcomes?
Since each coin can turn up in 2 ways, there are: outcomes.
Write . . . first half 's, the second half is.
. . .
. . .
Write . . . this time "change every two".
. . .
. . .
. . .
. . .
Write . . . this time "change every one" (alternate).
. . .
. . .
Write the three columns side-by-side:
. . .
. . . 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.