Basic Probability, part 4

August 23, 2009

All previous examples were with mutually exclusive events. These are events that do not have common cases or cannot occur at the same time. For example, the event “select a king” and the event “select an ace” from the deck are mutually exclusive, because we cannot select them both if we take one card.

The additional rule 1:
If two events (A and B) are mutually exclusive, then
P(A~or~B)=P(A)+P(B).

Thus the probability for the event “select a king or a queen” from a deck is
P(Queen~or~King)=P(Queen)+P(King)=\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}.

The example of not mutual exclusive events is “select an ace” and “select the spades” in a deck, because there are 4 aces, 13 spades cards, and there is one common case – the spades ace! The spades ace belongs to the group “aces” and to the group “spades” at the same time.

The additional rule 2:
If two events (A and B) are not mutually exclusive, then
P(A~or~B)=P(A)+P(B)-P(A~ and~ B).

In our case the probability of the event “select an ace or an spades card” is
P(Ace~or~Spades)=P(Ace)+P(Spades)-P(Ace~ and~ Spades ~at ~the~ same~ time)=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}.

Why do we need to subtract the probability of common cases P(A ~ and ~ B)?
Consider this issue in the next example. Look at the picture:
16

Let us find the probability of the event “randomly select a triangle or a yellow figure”. There are 12 figures at all and as we can see for our ask there are follow 7 figures:
33_2

So, the probability of the event “randomly select a triangle or a yellow figure” is P(Triangle~ or~ Yellow~figure)=\frac{7}{12}.

Good, but what do occur when we add the probability of the event “select a triangle” to the probability of the event “select a yellow figure”? There are 5 triangles and 5 yellow figures, so after add them up we get 10 figures, while we need 7 figures. The three excess figures occurs, because they are common and we have three yellow triangles, that belong to the group “triangles” and to the group “yellow figures” at the same time. Notice that after adding we count common figures two times: first, when we count triangles and second, when we count yellow figures:
33_3

Thus we need to subtract the number of common figures.

The probability of the event “randomly select a triangle or a yellow figure” is:
P(Triangle~ or~ Yellow~figure)=P(Triangle)+P(Yellow~figure)- \\-P(Triangle~ and~Yellow~figure)=\frac{5}{12}+\frac{5}{12}-\frac{3}{12}=\frac{7}{12}.

Advertisements