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This essential trains for: SAT-I, GMAT, AMC-8, Math Kangaroo 5-6, Math Kangaroo 7-8, Math Kangaroo 9-10.

Discrete probability refers to the probability of events that are countable (can be counted).

The set of all possible events is called sample space. For a discrete probability problem the sample space is countable and may have a finite or an infinite number of elements.

For example, the sample space of all the sums that can be obtained by throwing two fair dice is:

1 | 2 | 3 | 4 | 5 | 6 | |

1 | 2 | 3 | 4 | 5 | 6 | 7 |

2 | 3 | 4 | 5 | 6 | 7 | 8 |

3 | 4 | 5 | 6 | 7 | 8 | 9 |

4 | 5 | 6 | 7 | 8 | 9 | 10 |

5 | 6 | 7 | 8 | 9 | 10 | 11 |

6 | 7 | 8 | 9 | 10 | 11 | 12 |

The event "to throw a 4 with two dice" happens in 3 of the 36 possible sums. Its probability is:

The probability of an event is:

Obviously, such a problem always boils down to two counting problems. Sometimes, counting the size of the sample space first (quantity at the denominator), gives more insight into the problem.

Independent events

Assume the events a and b that are independent and happen with probabilities P(a) and P(b), respectively.

The event "a and b must both happen" has probability:

The event "either a or b must happen" has probability:

Therefore, many times, it is sufficient to break down the event of interest into simpler events that are connected by "and" and "or."

Example:

The seven Weasley siblings are to ride on five dragons. If the selection is random, what is the probability that Ron will end up riding with Bill? (characters from J.K. Rowling's "Harry Potter" series)

This complex event can be translated into a combination of simpler events: Ron chooses any dragon and Bill chooses the same dragon as Ron. There are two simple events connected by and.

The probability of the complex event is the product of the probabilities for the events: Ron chooses any dragon and, Bill chooses the same dragon as Ron.

The first event happens with probability 1 since Ron has to ride on some dragon.

The second event happens with probability 1/5 since there is only one dragon out of five that Bill may choose in a favorable case.

Therefore, the resulting probability is:

Let us now take a simple example and complicate it more and more. This will give us insight into how the problems are built.

All the examples are about a box containing 5 balls, labeled with numbers from 1 to 5. We pull balls out of the box blindfolded and we put them back in the box after having examined them (with replacement).

Example 1:

What is the probability of making two draws and getting the ball labeled "2" twice?

Translate the event into: pick the "2" and pick the "2". The individual probabilities have to be multiplied:

Example 2:

What is the probability of making a draw and getting the ball labeled "2" or the ball labeled "5"?

Translate the event into: pick the "2" or pick the "5". The individual probabilities must be added:

Example 3:

What is the probability of making two draws and getting the ball labeled "2" in the first draw and either the ball labeled "4" or the ball labeled "5" in the second draw?

Translate the event into: (pick the "2") and (pick the "5" or pick the "4"):

Now the same story only without replacement.

This situations are different because the sample spaces changes after each draw since the number of balls left in the box to pick from is smaller and smaller.

Example 4:

What is the probability of making two draws and getting the ball labeled "2" twice?

This event is clearly impossible, since after having picked the ball labeled "2" there is no other such ball left in the box. The probability of an impossible event is zero.

Example 5:

What is the probability of making two draws and getting the ball labeled "2" in the first draw and either the ball labeled "4" or the ball labeled "5" in the second draw?

The probabilities for the second draw have changed, since by now the size of the sample space has become 4:

Non-independent events

Example:

In the "Weasley siblings" experiment, what is the probability that Giny rides with Charlie or with Percy?

This event can be translated into: "Ginny chooses a dragon and Charlie chooses the same dragon or Ginny chooses a dragon and Percy chooses the same dragon." (Character names are from J.K. Rowling's "Harry Potter" series of youth novels.)

However, there is a catch: it is possible for Ginny, Charlie and Percy to choose the same dragon. This is an event for which we count the probability twice: once when we count the (Percy, Ginny) possibilities and once when we count the (Charlie, Ginny) possibilities.