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Statistics 101: Principles of Statistics11 chapters | 144 lessons | 9 flashcard sets

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Lesson Transcript

Instructor:
*Rudranath Beharrysingh*

In this lesson, we will look at generating a theoretical probability distribution for a discrete random variable and introduce the concept of expected value.

You may recall that a **probability distribution** summarizes the probability for a particular experiment.

Discrete probability distributions can be created from data or from theory based on what we know about a given experiment. Here we will look at theoretical discrete distributions like tossing coins, throwing dice, spinning a spinner or playing cards, but it could also be extended to other experiments where each outcome is known.

Suppose you toss two coins, and we want to know the various possible outcomes and the probabilities associated with them.

First, we need to look at all the possible outcomes. Well, when we toss two coins, we can get no heads, one head or two heads. Similarly we can get no tails, one tail or two tails. But note: if we got no heads, then we actually got two tails, and one head is the same as one tail, and two heads is equivalent to no tails. So, we have to decide if we want to look at the number of heads or the number of tails before we proceed.

For this problem, we'll just look at the number of heads. So, note the number of heads could be 0, 1 or 2, so we define our random variable X to be the number of heads when we toss two coins.

And we want to fill out a table like this:

Number of Heads When You Toss Two Coins | Probability of X: P(X) |
---|---|

0 | |

1 | |

2 |

But we need to calculate the probability for each value of X.

So, first of all, what is the probability that X is equal to 0? In shorthand math notation, we would write

*P(0) = ?*

Since *X = 0* means no heads (or two tails), and the probability of no head on one coin is 1/2, then the probability of no heads on both coins (no heads and no heads) is equal to *1/2 * 1/2 = 1/4*.

Note in probability, the word AND translates as MULTIPLY.

Thus, *P(0) = 1/4 or 0.25*

Number of Heads When You Toss Two Coins | Probability of X: P(X) |
---|---|

0 | 0.25 |

1 | |

2 |

Now let's calculate P(1): the probability of one head.

There are two possible outcomes. But, to look at this, let's label the coins as Coin One and Coin Two. So, the possible outcomes that have one head are: heads on Coin One and tails on Coin Two OR tails on Coin One and heads on Coin Two.

Let's look at the first outcome. The probability of heads on Coin One is 1/2, and then the probability of tails on Coin Two is 1/2. Multiplying them, we get 1/4. But we also have to look at the other outcome, which has exactly the same probability of 1/4. Since we are interested in only one head, we must take into consideration both of these outcomes, and so we add the possibilities (*1/4 + 1/4*) to get 1/2.

Note in probability, the word OR translates as ADD.

Thus the probability that X = 1, is 1/2 or in short hand math: *P(1) = 1/2 or 0.5*

Number of Heads When You Toss Two Coins | Probability of X: P(X) |
---|---|

0 | 0.25 |

1 | 0.5 |

2 |

Finally, we want to find the probability of getting two heads or P(2).

The probability of heads on each coin is 1/2, and so the probability of heads on Coin One and heads on Coin Two is *1/2 * 1/2 = 1/4*.

*P(2) = 1/4*.

We can summarize this in a table.

Number of Heads When You Toss Two Coins | Probability of X: P(X) |
---|---|

0 | 0.25 |

1 | 0.5 |

2 | 0.25 |

This is a probability distribution for the number of heads when you toss two coins.

A probability distribution contains all the values of the random variable X and its probability.

Note in a probability distribution, the probabilities sum to one.

Number of Heads When You Toss Two Coins | Probability of X: P(X) |
---|---|

0 | 0.25 |

1 | 0.5 |

2 | 0.25 |

Total = 1 |

Suppose you toss two coins and you want to know how many heads to expect. Well, since there are only two coins and the probability of heads is 50% (or 0.5), we would expect that when tossing two coins we would get one head. Thus, the expected value of X is 1 for this experiment.

But how do we calculate this, especially for larger distributions? Formally, **The Expected Value** is the sum of (X times P(X)) for all values of a random variable X, or *E(X) = The sum of X * P(X)*.

Applying this to the two coins example, we take each value of X and multiply by its probability and then sum them up. So we will have:

E(X) = (0 * 0.25) + (1 * 0.5) + (2 * 0.25)

Then, we sum these numbers up. So we have:

*0 + 0.5 + 0.5 = 1*

This is the expected value of X. In other words, we expect one head when we toss two coins. The calculation can be done using the probability distribution table:

Number of Heads When You Toss Two Coins | Probability of X: P(X) | X*P(X) |
---|---|---|

0 | 0.25 | 0 * 0.25 = 0 |

1 | 0.5 | 1 * 0.5 = 0.5 |

2 | 0.25 | 2 * 0.25 = 0.5 |

Sum: X * P(X) = 1 |

This does not mean that we will always get one head when we toss two coins. It means that, on average, when tossing two coins, you will get one head. Can you see where this information will be useful?

The expected value can also be applied to gambling and insurance. We'll look at gambling.

If you visit any casino, you are likely to see at least one table with a spinning wheel and ball. This game is quite popular and is called roulette.

In the game of roulette, you spin a wheel that contains a ball. There are different versions of the wheel, but let's suppose the wheel has 38 slots, and the ball can land on any one of 38 slots in the wheel. The slots are numbered 0, 00 and 1 through 36. The slots with 0 and 00 are colored green. The slots numbered 1 through 36 are colored black and red (18 numbers for each color).

Suppose the casino offers a payout of $10 if you bet $1 on green, and the ball lands on green. Remember, the green slots are the slots with 0 and 00. What can you expect to make or lose if you keep betting on green?

At first it may seem like a good deal, $10 for a $1 bet. But how often will the ball land on green?

To answer this question, we can set up a probability distribution for the amount of money you can win or lose per spin. Thus, we let X be the amount of money you make or lose per spin. What values can X take on, and is X discrete or continuous?

Well, if you lose on a bet, you lose just a $1, and if you win on a bet you win $10, but it cost you $1 to play, so your net gain is $9. So X can be two values, -1 for losing $1 and +9 for gaining $9. And since X can only equal these two distinct values, X is discrete.

X (Value of Game to You) | P(X) |
---|---|

-$1 | |

+$9 |

What are the probabilities associated with these values of X?

Remember, you only win if the ball lands on the green slots. Since there are only two green slots out of 38 slots, and they are all equally likely, the probability that the ball lands on a green spot is 2 out of 38 or 1/19.

Anything else, you lose. Unfortunately for you, there are 36 other slots, and so the probability you lose on a spin is 36 out of 38 or 18/19.

What can you expect to win if you keep betting $1 on green? Well there's a 1/19 probability you win $9 and an 18/19 probability you lose $1. So:

*9 * 1/19 + (-1 * 18/19) = -9/19 or -$0.47 per spin*

So you can expect to lose $0.47 on average per $1 bet on green. Suppose you bet $1 on green 100 times, you could expect to lose 0.47 * $100, which is equal to $47!

In this lesson, we looked at discrete probability distributions. For classical experiments, such as tossing coins, rolling dice, card games or games of chance like roulette, we can generate the distribution using what we know about the game and the rules of probability. We saw that the sum of the probabilities in a discrete distribution is 1. In addition, we learned how to calculate the expected value of the distribution by multiplying each value of the random variable X by its probability and then adding the results. **The expected value** is the sum of (X times P(X)) for all values of a random variable X. We also saw that the expected value of X has useful applications in gambling.

Following this lesson, you can:

- Chart the expected probability of an outcome.
- Calculate the expected value to determine the probability of an outcome.

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Statistics 101: Principles of Statistics11 chapters | 144 lessons | 9 flashcard sets

- Go to Probability

- Random Variables: Definition, Types & Examples 9:53
- Finding & Interpreting the Expected Value of a Discrete Random Variable 5:25
- Developing Discrete Probability Distributions Theoretically & Finding Expected Values 9:21
- Dice: Finding Expected Values of Games of Chance 13:36
- Blackjack: Finding Expected Values of Games of Chance with Cards 8:41
- Poker: Finding Expected Values of High Hands 9:38
- Poker: Finding Expected Values of Low Hands 8:38
- Lotteries: Finding Expected Values of Games of Chance 11:58
- Comparing Game Strategies Using Expected Values: Process & Examples 4:31
- How to Apply Discrete Probability Concepts to Problem Solving 7:35
- Binomial Experiments: Definition, Characteristics & Examples 4:46
- Finding Binomial Probabilities Using Formulas: Process & Examples 6:10
- Practice Problems for Finding Binomial Probabilities Using Formulas 7:15
- Finding Binomial Probabilities Using Tables 8:26
- Mean & Standard Deviation of a Binomial Random Variable: Formula & Example 6:34
- Solving Problems with Binomial Experiments: Steps & Example 5:03
- Go to Discrete Probability Distributions

- Go to Sampling

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