Question 1205515
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The probability distribution is
<table border = "1" cellpadding = "5"><tr><td># heads</td><td># tails</td><td>X = heads-tails</td><td>P(X)</td></tr><tr><td>4</td><td>0</td><td>4</td><td>1/16</td></tr><tr><td>3</td><td>1</td><td>2</td><td>4/16</td></tr><tr><td>2</td><td>2</td><td>0</td><td>6/16</td></tr><tr><td>1</td><td>3</td><td>-2</td><td>4/16</td></tr><tr><td>0</td><td>4</td><td>-4</td><td>1/16</td></tr></table>


Below I'll explain how I got each P(X) probability value.


There are n = 4 tosses of the coin.
That gives 2^n = 2^4 = 16 different outcomes.


Of those outcomes, there's only one way to get all heads. Same goes for all tails. That explains the 1/16 probability values for the first and last rows.


If there are 3 heads, then there are 4 ways to have this situation. This is because there are 4 places to put the tail. Those 4 outcomes are:<ol><li>HHHT</li><li>HHTH</li><li>HTHH</li><li>THHH</li></ol>Due to symmetry, the same idea applies if there are 3 tails.
So that's how we get a probability of 4/16 for the 2nd row and 2nd to last row.


If there are 2 heads, then there are 4C2 = 6 ways to arrange them. The 4C2 refers to the nCr combination formula. Such values can be found in Pascal's Triangle. The 6 ways to have 2 heads and 2 tails are listed here<ol><li>HHTT</li><li>HTTH</li><li>HTHT</li><li>TTHH</li><li>THTH</li><li>THHT</li></ol>So that's how I'm getting 6/16 for the probability of 2 heads.


I have not reduced the fractions in the P(X) column because I wanted to keep the denominators the same. But if you wanted you could reduce the fractions.
4/16 = 1/4
6/16 = 3/8


Or you can convert all the fractions to decimal form.
1/16 = 0.0625
4/16 = 1/4 = 0.25
6/16 = 3/8 = 0.375
Each decimal value is exact.


Two things to notice:<ul><li>Each P(X) value is between 0 and 1.</li><li>The P(X) values add to 1.</li></ul>
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