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13.2: Complements and Unions of Events

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Complementary Events

Now let us examine the probability that an event does not happen. As in the previous section, consider the situation of rolling a six-sided die and first compute the probability of rolling a six: the answer is P(six)=16. Now consider the probability that we do not roll a six: there are 5 outcomes that are not a six, so the answer is P(not a six)=56. Notice that P(six)+P(not a six)=16+56=66=1

This is not a coincidence. Consider a generic situation with n possible outcomes and an event E that corresponds to m of these outcomes. Then the remaining nm outcomes correspond to E not happening, thus P(not E)=nmn=nnmn=1mn=1P(E)

Complement of an Event

The complement of an event E is the event “E doesn’t happen”

The notation ˉE is used for the complement of event E.

We can compute the probability of the complement using P(ˉE)=1P(E)

Notice also that P(E)=1P(ˉE)

Example 1

If you pull a random card from a deck of playing cards, what is the probability it is not a heart?

Solution

There are 13 hearts in the deck, so

P(heart)=1352=14.

The probability of not drawing a heart is the complement:

P( not heart )=1P( heart )=114=34

Example 2

Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin or a 6 on the die.

Solution

Here, there are still 12 possible outcomes:

{H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}

By simply counting, we can see that 7 of the outcomes have a head on the coin or a 6 on the die or both – we use or inclusively here (these 7 outcomes are H1,H2,H3,H4,H5,H6,T6), so the probability is 712. How could we have found this from the individual probabilities?

As we would expect, 12 of these outcomes have a head, and 16 of these outcomes have a 6 on the die. If we add these, 12+16=612+212=812, which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head and rolling a 6 is 112.

If we subtract out this double count, we have the correct probability: 812112=712.

P(A or B)

The probability of either A or B occurring (or both) is P(A or B)=P(A)+P(B)P(A and B)

Example 3

Suppose we draw one card from a standard deck. What is the probability that we get a Queen or a King?

Solution

There are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is:

P( King or Queen)=852

Note that in this case, there are no cards that are both a Queen and a King, so P(King and Queen)=0. Using our probability rule, we could have said:

P( King or Queen)=P( King )+P( Queen)P( King and Queen)=452+4520=852

In the last example, the events were mutually exclusive, so P(A or B)=P(A)+P(B).

Mutually Exclusive

Two events A and B are mutually exclusive if they have no outcomes in common.

Thus, P(A or B)=P(A)+P(B).

Example 4

Suppose we draw one card from a standard deck. What is the probability that we get a red card or a King?

Solution

  • Half the cards are red, so P(Red)=2652
  • There are four kings, so P(King)=452
  • There are two red kings, so P(Red and King)=252

We can then calculate

P(Red or King)=P(Red)+P(King)P(Red and King)=2652+452252=2852=713

Try it Now 1

In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you reach in and randomly grab a pair of socks and a tee shirt, what the probability you pulled either a pair of white socks or a white tee shirt?

Answer

P( white sock )=610

P( white tee )=37

P( white sock and white tee )=61037=935

P( white sock or white tee )=610+37935=2735


This page titled 13.2: Complements and Unions of Events is shared under a CC BY-SA license and was authored, remixed, and/or curated by David Lippman (The OpenTextBookStore) .

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