## Probability - Statistics 1

P(A ∩ B) = P(A) x P(B) will be true if they are independent.

**- The probability is not affected by the outcome of the previous event. Examples: flipping a coin or rolling a dice.**__Independent events__P(A ∩ B) = P(A) x P(B) will be true if they are independent.

**- The probability is affected by the outcome of the previous event. Example: picking an item from a bag without replacement. The probability of picking each item the second time round will depend on which item was taken out the first time.**__Dependent events__

**– the events cannot both be true or they can’t happen at the same time.**__Mutually exclusive events__P(A ∩ B) = 0

P(A ∪ B) = P(A) + P(B)

P(A ∪ B) = P(A) + P(B)

**– the events can both be true or they can happen at the same time.**__Non-mutually exclusive events__P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

If several events happen one after the other then you need to work out the probability for each of events and you need to multiply them together. To calculate the probability of successive events you can draw a tree diagram.

For example, flipping a coin:

P(Heads) = ½

P(Two heads in a row) = ½ x ½ = ¼

P(Three heads in a row) = ½ x ½ x ½ = 1/8

If there is an event where there is more than one way in which the event can happen, then you need to work out the probability for all of the different cases and you need to add them together.

Example: There are 4 blue marbles and 6 green marbles in a bag. Two marbles are taken out at random. Find the probability two different coloured marbles are picked.

__Probability of Successive Events__If several events happen one after the other then you need to work out the probability for each of events and you need to multiply them together. To calculate the probability of successive events you can draw a tree diagram.

For example, flipping a coin:

P(Heads) = ½

P(Two heads in a row) = ½ x ½ = ¼

P(Three heads in a row) = ½ x ½ x ½ = 1/8

__Probability of events with several possible cases__If there is an event where there is more than one way in which the event can happen, then you need to work out the probability for all of the different cases and you need to add them together.

Example: There are 4 blue marbles and 6 green marbles in a bag. Two marbles are taken out at random. Find the probability two different coloured marbles are picked.

Cases |
Probability |

Blue and Green |
4/10 x 6/9 (successive events so you multiply) |

Green and Blue |
6/10 x 4/9 (successive events so you multiply) |

P(Different coloured marbles = 4/10 x 6/9 + 6/10 x 4/9 = 8/15 (Different cases for the same event of picking different coloured marbles, so you add).

__Venn Diagram Examples__‘∩’ means "and". For ‘∩’ both events need to be true.

‘∪’ means "or". For ‘∪’ it needs to be true for

A’ means "not A". P(A’ ∩ B) means the probability of

‘∪’ means "or". For ‘∪’ it needs to be true for

**of the events.**__at least one__A’ means "not A". P(A’ ∩ B) means the probability of

*not A*and*B*happening.P(A) = 2 + 3 |
P(A ∩ B’) = 2 |

P(B) = 3 + 4 |
P(A’ ∪ B) = 1 + 3 + 4 |

P(A ∩ B) = 3 |
P(A ∪ B’) = 1 + 2 + 3 |

P(A ∪ B) = 2 + 3 + 4 |
P(A’ ∪ B’) = 1 |

P(A’ ∩ B’) = 1 |
P(A ∩ B)’ = 1 + 2 + 4 |

P(A’ ∩ B) = 4 |
P(A ∪ B)’ = 1 |

(N.B. all probabilities are less than 1 - probabilities in above table are all divided by 10 because there are 10 total in the set.)

__Conditional Probability__The probability of A happening given that B has happened = Probability of A and B happening divided by the probability of the given event B.