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What is the probability that in a family with two children, both are boys? What is the probability that in a family with two children, both are boys, if we already know that one is a boy? What changes if we know that the older child is a boy? What makes the second and third probability questions different from the first? (Assume that the probability of having a boy is 1/2, and the sex of each child is independent from the sex of other children.)
Consider the Venn diagram for two events E and F shown in the figure We seek . Let n(E), n(F), and be the number of ways that events E, F and can occur among all the outcomes of the experiment. Now knowing the additional information that event F has occurred, we find by focusing our attention on the region in the event circle labeled F. We evaluate the number
which represents the number of times that E has occurred among all those outcomes in which F has occurred. Now
Sometimes it is easy to calculate the conditional probability, but may be hard or confusing to compute the probability of the intersection. In such a case, we can turn the conditional probability formula around by multiplying through by the denominator and obtaining:
Suppose I have rolled two dice and I announce that the sum of the two dice is 5 or less. Compute the probability that the sum of the two dice is an even number.
Solution: First I list all the possible to ways to roll two dice and have a sum less than or equal to 5:
Dice 1 | Dice 2 | Sum |
1 | 1 | 2 |
1 | 2 | 3 |
2 | 1 | 3 |
1 | 3 | 4 |
3 | 1 | 4 |
2 | 2 | 4 |
1 | 4 | 5 |
4 | 1 | 5 |
2 | 3 | 5 |
3 | 2 | 5 |
We see that there are 10 possible outcomes. We are interested in how many ways to roll to get an even number in the sum column. Observe that 2 and 4 are the only even numbers less than 5. There is one way to roll the dice and have their sum be 2. There are three ways to roll the dice and have their sum be 4. Hence, the probability is .
Suppose an urn contains 8 red balls and 4 white balls. We draw 2 balls from the urn without replacement. If we assume that as we draw the each ball in the urn is equally likely to be drawn, what is the probability that both balls drawn are red?
Let R_{1} and R_{2} denote respectively the events that the first and the second ball drawn is red. Now given that the first ball selected is red, there 7 remaining red balls and 4 white balls and so . As , the desired probability is
Note that another way to work this problem is use combinations by thinking of drawing a pair of balls in quick succession:
An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the first 2 selected are white and the last 2 selected are black?
Let the event that the first ball drawn is white be W_{1}, the second ball selected is white be W_{2}, the third ball selected is black be B_{3} and the fourth ball selected is black is B_{4}. The we are interested in the probability of Now
But further decomposing
and as in the previous example:
Putting them altogether:
Fortunately, each conditional probability is easy to compute:
This section is adapted from: Finite Mathematics, by Karl J. Smith, Scott, Foresman and Company, Glenview, IL, 1975. Some of the problems and examples are from A First Course in Probability, by Sheldon Ross.
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