7.2: Conditional Probability

Often it is required to compute the probability of an event given that another event has occurred.

Example \(\PageIndex{1}\)

What is the probability that two cards drawn at random from a deck of playing cards will both be aces?

Solution

It might seem that you could use the formula for the probability of two independent events and simply multiply \(\dfrac{4}{52} \cdot \dfrac{4}{52}=\dfrac{1}{169}\). This would be incorrect, however, because the two events are not independent. Once the first card is drawn, there are only 51 cards remaining in the deck. If the first card drawn is an ace, then the probability that the second card is also an ace would be \(\dfrac{3}{51}\) because there would only be three aces left in the deck.

Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the conditional probability of drawing an ace. In this case, the "condition" is that the first card is an ace. Symbolically, we write this as:

\(P(\) ace on second draw \(\mid\) an ace on the first draw).

The vertical bar "|" is read as "given," so the above expression is short for "The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw." What is this probability? After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is \(\dfrac{3}{51}=\dfrac{1}{17}\).

Thus, the probability of both cards being aces is \(\dfrac{4}{52} \cdot \dfrac{3}{51}=\dfrac{12}{2652}=\dfrac{1}{221}\).

Definition: Conditional Probability

The probability that event B occurs, given that event A has happened, is represented as P(B | A) This is read as “the probability of B given A”

Example \(\PageIndex{2}\)

Find the probability that a die rolled shows a 6, given that a flipped coin shows a head.

Solution

These are two independent events, so the probability of the die rolling a 6 is \(\dfrac{1}{6}\), regardless of the result of the coin flip.

Example \(\PageIndex{3}\)

The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their cars. Find the probability that a randomly chosen person:

a) Has a speeding ticket given they have a red car
b) Has a red car given they have a speeding ticket

	Speeding Ticket	No Speeding Ticket	Total
Red car	15	135	150
Not red car	45	470	515
Total	60	605	665

Solution

a) Since we know the person has a red car, we are only considering the 150 people in the first row of the table. Of those, 15 have a speeding ticket, so

\[
\mathrm{P}(\text { ticket } \mid \text { red car })=\dfrac{15}{150}=\dfrac{1}{10}=0.1
\]

b) Since we know the person has a speeding ticket, we are only considering the 60 people in the first column of the table. Of those, 15 have a red car, so

\[
\mathrm{P}(\text { red car } \mid \text { ticket })=\dfrac{15}{60}=\dfrac{1}{4}=0.25 \text {. }
\]

Notice from the last example that P(B | A) is not equal to P(A | B).

These kinds of conditional probabilities are what insurance companies use to determine your insurance rates. They look at the conditional probability of you having an accident,
given your age, your car, your car color, your driving history, etc., and price your policy based on that likelihood.

P(A and B) for dependent events

If Events A and B are dependent (not independent), then P(A and B) = P(A) · P(B | A)

Example \(\PageIndex{4}\)

If you pull 2 cards out of a deck, what is the probability that both are spades?

Solution

The probability that the first card is a spade is \(\dfrac{13}{52}\).

The probability that the second card is a spade, given the first was a spade, is \(\dfrac{12}{51}\), since there is one less spade in the deck, and one less total cards.

The probability that both cards are spades is \(\dfrac{13}{52} \cdot \dfrac{12}{51}=\dfrac{156}{2652} \approx 0.0588\)

You Try It \(\PageIndex{1}\)

If you draw 2 cards from a standard deck, what is the probability that both cards are red
kings?

Answer

0.00075

Example \(\PageIndex{5}\)

A home pregnancy test was given to women, then pregnancy was verified through blood tests. The following table shows the home pregnancy test results. Find

a) P(not pregnant | positive test result)
b) P(positive test result | not pregnant)

	Positive test	Negative test	Total
Pregnant	70	4	74
Not Pregnant	5	14	19
Total	75	18	93

Solution

a) Since we know the test result was positive, we're limited to the 75 women in the first column, of which 5 were not pregnant.

\[P(\) not pregnant \(\mid\) positive test result \()=\dfrac{5}{75} \approx 0.067\].

b) Since we know the woman is not pregnant, we are limited to the 19 women in the second row, of which 5 had a positive test.

\[
P \text { (positive test result } \mid \text { not pregnant })=\dfrac{5}{19} \approx 0.263
\]

The second result is what is usually called a false positive: A positive result when the woman is not actually pregnant.