2 Conditional probability - dependent events

Suppose a bag contains 6 balls, 3 red and 3 white. Two balls are chosen (without replacement) at random, one after the other. Consider the two events $R,W$ :

$R$ is event “first ball chosen is red”

$W$ is event “second ball chosen is white”

We easily find $\text{P}\left(R\right)=\frac{3}{6}=\frac{1}{2}$ . However, determining the probability of $W$ is not quite so straightforward. If the first ball chosen is red then the bag subsequently contains 2 red balls and 3 white. In this case $\text{P}\left(W\right)=\frac{3}{5}$ . However, if the first ball chosen is white then the bag subsequently contains 3 red balls and 2 white. In this case $\text{P}\left(W\right)=\frac{2}{5}$ . What this example shows is that the probability that $W$ occurs is clearly dependent upon whether or not the event $R$ has occurred. The probability of $W$ occurring is conditional on the occurrence or otherwise of $R$ .

The conditional probability of an event $B$ occurring given that event $A$ has occurred is written $\text{P}\left(B|A\right)$ . In this particular example

$\text{P}\left(W|R\right)=\frac{3}{5}\text{and}\text{P}\left(W|R^{\prime }\right)=\frac{2}{5}.$

Consider, more generally, the performance of an experiment in which the outcome is a member of an event $A$ . We can therefore say that the event $A$ has occurred. What is the probability that $B$ then occurs? That is what is $\text{P}\left(B|A\right)$ ? In a sense we have a new sample space which is the event $A$ . For $B$ to occur some of its members must also be members of event $A$ . So, for example, in an equi-probable space, $\text{P}\left(B|A\right)$ must be the number of outcomes in $A\cap B$ divided by the number of outcomes in $A$ . That is

$\text{P}\left(B|A\right)=\frac{\text{number of outcomes in}A\cap B}{\text{number of outcomes in }A}.$

Now if we divide both the top and bottom of this fraction by the total number of outcomes of the experiment we obtain an expression for the conditional probability of $B$ occurring given that $A$ has occurred:

To illustrate the use of conditional probability concepts we return to the example of the bag containing 3 red and 3 white balls in which we consider two events:

• $R$ is event “first ball is red”
• $W$ is event “second ball is white”

Let the red balls be numbered 1 to 3 and the white balls 4 to 6. If, for example, $\left(3,5\right)$ represents the fact that the first ball is 3 (red) and the second ball is 5 (white) then we see that there are $6\times 5=30$ possible outcomes to the experiment (no ball can be selected twice).

If the first ball is red then only the fifteen outcomes $\left(1,x\right),\left(2,y\right),\left(3,z\right)$ are then possible (here $x\ne 1$ , $y\ne 2$ and $z\ne 3$ ). Of these fifteen, the six outcomes $\left\{\left(1,2\right),\left(1,3\right),\left(2,1\right),\left(2,3\right),\left(3,1\right),\left(3,2\right)\right\}$ will produce the required result, i.e. the event in which both balls chosen are red, giving a probability: $\text{P}\left(B|A\right)=\frac{6}{15}=\frac{2}{5}$ .

Example 11

A box contains six 10 $\Omega$ resistors and ten 30 $\Omega$ resistors. The resistors are all unmarked and are of the same physical size.

1. One resistor is picked at random from the box; find the probability that:
   1. It is a 10 $\Omega$ resistor.
   2. It is a 30 $\Omega$ resistor.
2. At the start, two resistors are selected from the box. Find the probability that:
   1. Both are 10 $\Omega$ resistors.
   2. The first is a 10 $\Omega$ resistor and the second is a 30 $\Omega$ resistor.
   3. Both are 30 $\Omega$ resistors.

Solution

1. 1. As there are six 10 $\Omega$ resistors in the box that contains a total of 6 + 10 = 16 resistors, and there is an equally likely chance of any resistor being selected, then

      $\text{P}\left(10\Omega \right)=\frac{6}{16}=\frac{3}{8}$

   2. As there are ten 30 $\Omega$ resistors in the box that contains a total of 6 + 10 = 16 resistors, and there is an equally likely chance of any resistor being selected, then

      $\text{P}$ (30 $\Omega$ ) $=\frac{10}{16}=\frac{5}{8}$

2. 1. As there are six 10 $\Omega$ resistors in the box that contains a total of 6 + 10 = 16 resistors, and there is an equally likely chance of any resistor being selected, then

      $\text{P}$ (first selected is a 10 $\Omega$ resistor) $=\frac{6}{16}=\frac{3}{8}$

      If the first resistor selected was a 10 $\Omega$ one, then when the second resistor is selected, there are only five 10 $\Omega$ resistors left in the box which now contains 5 + 10 = 15 resistors.

      Hence, $\text{P}$ (second selected is also a 10 $\Omega$ resistor) $=\frac{5}{15}=\frac{1}{3}$

      And, $\text{P}$ (both are 10 $\Omega$ resistors) $=\frac{3}{8}\times \frac{1}{3}=\frac{1}{8}$

   2. As before, $\text{P}$ (first selected is a 10 $\Omega$ resistor) $=\frac{6}{18}=\frac{3}{8}$

      If the first resistor selected was a 10 $\Omega$ one, then when the second resistor is selected, there are

      still ten 30 $\Omega$ resistors left in the box which now contains 5 + 10 = 15 resistors. Hence,

      $\text{P}$ (second selected is a 30 $\Omega$ resistor) $=\frac{10}{15}=\frac{2}{3}$

      And, $\text{P}$ (first was a 10 $\Omega$ resistor and second was a 30 $\Omega$ resistor) $=\frac{3}{8}\times \frac{2}{3}=\frac{1}{4}$

   3. As there are ten 30 $\Omega$ resistors in the box that contains a total of 6 + 10 = 16 resistors,

      and there is an equally likely chance of any resistor being selected, then

      $\text{P}$ (first selected is a 30 $\Omega$ resistor) $=\frac{10}{16}\times \frac{5}{8}$

      If the first resistor selected was a 30 $\Omega$ one, then when the second resistor is selected,

      there are only nine 30 $\Omega$ resistors left in the box which now contains 5 + 10 = 15 resistors.

      Hence, $\text{P}$ (second selected is also a 30 $\Omega$ resistor) $=\frac{9}{15}=\frac{3}{5}$

      And, $\text{P}$ (both are 30 $\Omega$ resistors) $=\frac{5}{8}\times \frac{3}{5}=\frac{3}{8}$