We use binomial theorem in tandem with probabilities to get our probability distributions. A Binomial distribution is used to model specific independant trials in which the outcome is either success or failure and the probability of success is the same each trial. Probability is how we divide 1 by into cases.
Bernoulli Trials
To find out if a problem involves binomial distribution, you consult Bernoulli Trials
Probability in a Binomial Distribution
P(x) = p is the probability of success on any individual trial. (eg. 50% a coin toss) q = 1 - p n = number of trials P total number of trials x = result
Expectation Formula
E(X) = np
Examples
Example 1
heads probability 50%, tails probability 50%. probability distribution of flipping 3 times Note that h + t = 1. 100% divided up into probability sets
Solution
= (1/8) + (3/8) + (3/8) + (1/8) first term is P(no tails), P(no tails) = 1/8 P(one tail) = 2nd term. P(one tails) = 3/8 P(second tail) = 3rd term. P(second tails) = 3/8 P(third tail) = 1/8
Example 2
Flip a coin 4 times. GIve the probability distribution where ‘T’ is the successful outcome T = the number of Tails. What is the expected number of ‘T’s?
| T = t | P(T = 1) |
|---|---|
| 0 | 1/16 |
| 1 | = |
| 2 | = |
| 3 | 4/16 |
| 4 | 1/16 |
| This follows pascal triangle you see! If you add up all the probabilities, they should equal 1. | |
| 1/16 + 4/16 + 6/16 + 4/16 + 1/16 = 16/16 = 1! | |
| The expected number is 2. |
Example 3
Game consists of rolling a pair of dice 10 times. for each sum that equals 6,7,8 or 2, you win 1 dollar. it costs 5 dollars to play the game. Is this a fair game?
Solution
n = 10 p = … cases for 7 : {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} = 6 cases cases for 6 : {(1,5),(2,4),(3,3),(4,2),(5,1)} = 5 cases cases for 8 : {(2,6),(3,5),(4,4),(5,3),{6,2}} = 5 cases 6 + 5 + 5 total possibilities = 6^2 = 36 p = 16/36
E(X) = np = 10 (4/9) = 40/9 = 4.44
4.44 is smaller than 5 dollar. not fair!