We'll build this up with an example. If I flip the coin once there are two possible outcomes: H and T. Each outcome occurs 1 time.

If I flip a coin twice there are 4 possible outcomes HH, TH, HT and TT. Notice that the outcome of 1 head and 1 tail occurs twice (or two different ways). Putting this into a table:

**Flips** |
**Outcomes** |

1 |
H,T |

2 |
HH, TH, HT,TT |

3 |
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT |

For three flips of a coin there are 4 separate outcomes 3 Heads (1 occurrence), 2 Heads and 1 Tail (3 occurrences), 1 Head and 2 Tail (3 occurrences) and 3 Tails (1 occurrence).

Now if we make a new table with the number of occurrences of each outcome we get something like:

**Flips** |
**Occurrences** |

1 |
1 1 |

2 |
1 2 1 |

3 |
1 3 3 1 |

This pattern should look familiar. Its the 2nd through 4th row of Pascals triangle, which is the pattern that binomial expansions follow! Lets expand the table bigger to try to explain a bit more.

**Flips** |
**Occurrences** |

1 |
1 1 |

2 |
1 2 1 |

3 |
1 3 3 1 |

4 |
1 4 6 4 1 |

5 |
1 5 10 10 5 1 |

If we look at the last row this represents flipping a coin 5 times. This shows us that there are only 1 way of getting 5 heads, 5 ways of getting 4 heads and 1 tails, 10 ways of getting 3 heads and 2 tails, etc…

This is the very same idea as with binomial expansions!

##### Example 1

If we want to know the probability of a certain outcome, say 4 heads out of 5 flips, first we need to calculate the probability this occurring:

(3)
\begin{align} \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{32} \end{align}

BUT! This can happen 5 different ways for we must multiply by 5 giving a final probability of: $\frac{5}{32}$.

Using the formula from above:

(4)
\begin{align} P(X=4)= \left ( \begin{matrix}5\\ 4\end{matrix}\right ) (1-0.5)^{5-4}(0.5)^4 \end{align}

##### Example 2

Find the probability of rolling a number less than 5 on a die for 3 out of 4 rolls.

First the probability of rolling less than a 5 is $\frac{4}{6}$ and the probability of rolling 5 or more is $\frac {2}{6}$. So the total probability is:

(5)
\begin{align} \frac{4}{6} \cdot \frac{4}{6} \cdot \frac{4}{6} \cdot \frac{2}{6} = \frac{128}{1296} \end{align}

But, there are 4 different ways for this to occur (SSSF, SSFS, SFSS, FSSS - S is for success and F is for failure) so we have to multiply our probability by 4.

(6)
\begin{align} P(X=3)=\frac{512}{1296} \end{align}

Using the equation from above:

(7)
\begin{align} P(X=3)= \left ( \begin{matrix}4\\ 3\end{matrix}\right ) (1-\frac{2}{6})^{4-3}(\frac{4}{6})^3 \end{align}

#### Okay, but the 1st Row of Pascal's Triangle is Missing!

This corresponds to $n=0$ and $r=0$. Which in the language of probability means we did zero trials ($n=0$) to which there is only 1 outcome (the value in the 1st row) and that is zero successes ($r=0$).

Why is the probability of a successful event 0.5? If you want the probability of getting 3 heads when flipping the coin 5 times, isn't the probability of success 3 out of 5 (which is 0.6)?

ReplyOptionsThe "successful event" is getting heads once so P(heads)= 0.5 Another way to think of this is each event is separate or independent, i.e. each flip of the coin does not affect the next.

In order to find the probability of getting EXACTLY 3 heads out of 5 we need to consider all the different ways it can be done.

HHHTT

HHTHT

HHTTH

etc…

The coefficient on equation (1) gives us the number of ways it can happen dependent on

nandr.Note that the probability of HHHTT is exactly the same as HHTHT (or for that matter HHHHH). That is P(HHHTT) = (0.5)(0.5)(0.5)(0.5)(0.5)=0.031250 However, there is only one way to get HHHHH which makes it "rare" whereas there are many ways to get 3 heads out of 5.

Hope that helps.

ReplyOptionsthe probability of a successful event is the probability of getting heads, which on a two sided coin, is one in two —> 0.5

the 3 times makes no relevance to the probability of getting a head, since doing it three times doesn't affect the outcome of the coin being heads or tails

ReplyOptionsLine marked (7) above is incorrect. Probability of success and failures are swapped in the formula

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