## S7 Conditional probability

### Dependent Events

Two events are dependent if the outcome or occurrence of the first affects the outcome or occurrence of the second so that the probability is changed.

#### Example

A card is chosen at random from a pack. If the first card chosen is the jack of diamonds and it is **not replaced** what is the probability that the second card is

- a diamond? 1Because the first card was the jack of diamonds and not replaced, there is one less diamond in the pack and one less card in the pack.

Pr(\(\diamondsuit\)) = \(\dfrac{12}{51}\) = \(\dfrac{4}{17}\).

- a jack?

Pr(jack) = \(\dfrac{3}{51}\) = \(\dfrac{1}{17}\)

- the queen of clubs?

Pr(Q\(\clubsuit\)) = \(\dfrac{1}{51}\)

The events J\(_{1}\) “jack of diamonds on the first draw” and D\(_{2}\) “a diamond on the second draw” are dependent when there is no replacement.

The probability of choosing a diamond on the second draw given that the jack of diamonds was chosen on the first pick is called a conditional probability.

We say “The probability of D\(_{2}\) given J\(_{1}\) is \(\dfrac{4}{17}\)” and write this as \[\begin{align*} \Pr\left(\text{D$_{2}$ |$\text{J$_{1}$ }$ }\right) & =\frac{4}{17}. \end{align*}\]

### Multiplication Rule

When two events, \(A\) and \(B\), are dependent, the probability of both occurring is: \[ Pr(A\:and\:B)=Pr(A\cap B)=Pr(A)\times Pr(B|A). \]

#### Example

Find the probability of obtaining two jacks if two cards are drawn is succession from a pack

with replacement

without replacement.

Solution a).

If the cards are replaced then the events are independent and

\[\begin{align*} Pr(J_{1}\cap J_{2}) & =Pr(J_{1})\times Pr(J_{2})\\ & =\frac{4}{52}\times\frac{4}{52}\\ & =\frac{1}{169}. \end{align*}\]

Solution b).

If the cards are not replaced then the probability of the second draw depends on the first draw: \[\begin{align*} Pr(J_{1}\cap J_{2}) & =Pr(J_{1})\times Pr(J_{2}|J_{1})\\ & =\frac{4}{52}\times\frac{3}{51}\\ & =\frac{1}{221}. \end{align*}\]

### Conditional Probability

The multiplication rule for dependent events can be rearranged to find a conditional probability: \[\begin{align*} Pr(B|A) & =\frac{Pr(A\cap B)}{Pr(A)} & \left(1\right)\\ or\\ Pr(A|B) & =\frac{Pr(A\cap B)}{Pr(B)}. & \left(2\right) \end{align*}\]

#### Examples

**1. **Find the \(Pr(A|B)\) if \(Pr(A)=0.7,Pr(B)=0.5\) and \(Pr(A\cup B)\) \(=0.8\).

Solution:

First find \(Pr(A\cap B)\). 2 We need this to use equation \(\left(2\right).\)

\[\begin{align*} Pr(A\cup B) & =Pr(A)+Pr(B)-Pr(A\cap B)\\ 0.8 & =0.7+0.5-Pr(A\cap B)\\ Pr(A\cap B) & =0.7+0.5-0.8\\ & =0.4. \end{align*}\]

and

\[\begin{align*} Pr(A|B) & =\dfrac{Pr(A\cap B)}{Pr(B)}\\ & =\dfrac{0.4}{0.5}\\ & =0.8. \end{align*}\]

**2. **In a class of \(15\) boys and \(12\) girls two students are to be randomly chosen to collect homework.

What is the probability that both students chosen are boys?

Solution:

We have, 3 Note that the probability of a boy on the second choice, given a boy was chosen first, is \[\begin{align*} \Pr\left(B_{2}|B_{1}\right) & =\frac{14}{26} \end{align*}\] because there is one less boy and one less student to select.

\[\begin{align*} Pr(B1\cap B2) & =Pr(B_{1})\times Pr(B2|B1)\\ & =\dfrac{15}{27}\times\dfrac{14}{26}\\ & =\dfrac{210}{702}\\ & =\dfrac{35}{117}. \end{align*}\]

Another way to do conditional probability problems is to reduce the sample space, as in the example below.

**3. **Given the information in the following table find the probability that someone was sunburnt **given that they were not wearing a hat**.

Sunburnt | Not sunburnt | ||
---|---|---|---|

Hat | 3 | 77 | 80 |

No hat | 12 | 8 | 20 |

15 | 85 | 100 |

Solution:

Let \(H'\) be the event “the person is not wearing a hat”. We want \(\Pr\left(S|H'\right)\). The total sample space involves one hundred people. We can reduce this by confining out attention to those not wearing a hat. From row three of the table we see that of the 20 people not wearing a hat, 12 of them were sunburnt. So \[\begin{align*} \Pr\left(S|H'\right) & =\frac{12}{20}\\ & =\frac{3}{5}. \end{align*}\]

### Exercises

**1. **The results of a survey of music preferences are displayed in the Venn diagram below. 4 Image from Passy’s world of mathematics. http://passyworldofmathematics.com/three-circle-venn-diagrams/

Find the probability that a student likes rock music given that they like dance music.

\(\dfrac{5+8}{14+20+8+5}=\dfrac{13}{47}\).

**2. **Three cards are chosen at random from a pack without replacement. What is the probability of choosing 3 aces?

\(\dfrac{4}{52}\times\dfrac{3}{51}\times\dfrac{2}{50}=\dfrac{1}{5525}\) .

**3. **In a maths class of 20 students 5 failed the final exam. If two students are chosen at random without replacement, what is the probability that the first passed but the second failed?

\(\dfrac{15}{20}\times\dfrac{5}{19}=\dfrac{15}{76}\)

**4. **If \(Pr(X)=0.5,Pr(Y)=0.5\) and \(Pr(X\cap Y)=0.2\), find the probability of

\(Pr(X|Y)\)

\(Pr(X\cup Y)\)

\(Pr(X)\times Pr(Y|X)\)

\((a)\,0.4\quad(b)\,0.8\quad(c)\,0.2\)

**5. **In a three child family what is the probability that all three children will be girls given that the first child is a girl.

\(0.25\)

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