Learning Lab

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

1. a diamond? 11Because 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(\(\diamondsuit\)) = \(\dfrac{12}{51}\) = \(\dfrac{4}{17}\).

2. a jack?

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

3. 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 \[\begin{align*} \Pr\left(\text{D$_{2}$ |$\text{J$_{1}$ }$ }\right) & =\frac{4}{17}. \end{align*}\]

Example

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

1. with replacement

2. 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: \[\begin{align*} Pr(J_{1}\cap J_{2}) & =Pr(J_{1})\times Pr(J_{2}|J_{1})\\ & =\frac{4}{52}\times\frac{3}{51}\\ & =\frac{1}{221}. \end{align*}\]

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 \(Pr(A\cap B)\). 22 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, 33 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.

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 \[\begin{align*} \Pr\left(S|H'\right) & =\frac{12}{20}\\ & =\frac{3}{5}. \end{align*}\]