Mixed Questions on Probability

When we attempt to answer questions on any mathematical topic, we need to be able to choose, from among the techniques we have learned, the correct method for the task at hand. This chapter reviews some of the ideas previously presented concerning Probability.

Relative frequency

We often understand probability by considering an experiment of some kind that can be repeated many times and which has possible outcomes that are well defined. We interpret the proportion of occurrences of a particular outcome in all the trials of the experiment as the probability of that outcome.

Experiment and observation

Usually, we discover the relative frequencies of the outcomes of an experiment by actually performing the experiment with a sufficiently large number of trials. Sometimes, we cannot perform an experiment but must rely instead on repeated observations of many similar situations.

In experiments in the physical sciences, for example, it is usual to repeat a procedure many times and to note the number of numerical outcomes that fall within each of several measurement categories. The result is then given as an average of the outcomes together with a probability that the true value is somewhere within a particular interval around the mean value of the observed outcomes.

In medical research, it may be unethical or impractical to perform an experiment. Instead, observational studies are made, comparing outcomes over a large number of subjects who are similar in respects that are not thought to affect the results of the study but who are given different treatments or who have some other difference thought to be important. Again, the relative frequency of the different outcomes in the study are interpreted as probabilities.

Similarly, surveys conducted to ascertain product preferences or political opinions are observational studies. The individuals interviewed in the survey are the experimental subjects and the relative frequencies of the various possible responses are interpreted as probabilities.

Counting

Sometimes, it is possible to find a relative frequency and, hence, a probability by counting theoretically the number of ways in which each member of a set of possible outcomes can occur. Probabilities concerning card and other games are often of this type. The idea of counting the numbers of outcomes that fall into each of several categories leads to the mathematical field of combinatorics.

Examples

Example 1

A single card is dealt from a standard deck of 52 playing cards. What is the probability that the card will be an ace, a king, a queen or a jack?

We could perform this experiment, say, $50$50 times in the classroom to discover the relative frequency of the required event. (Note that the occurrence of any one of the four listed outcomes would mean the event has occurred.) A more accurate estimate of the relative frequency would be obtained if we performed the experiment, say, $1000000$1000000 times with a computer simulation.

Even better would be a counting argument. Since $16$16 out of the $52$52 cards are either an ace, a king, a queen or a jack, we see immediately that the relative frequency of the event in which we are interested is $\frac{16}{52}=\frac{4}{13}$1652​=413​. Therefore, the probability of this event is $\frac{4}{13}$413​

Tree Diagrams

Visual representations of an experimental situation can help in counting the possible outcomes or their relative frequencies. A tree diagram may clarify complicated multi-stage problems. 

Example 2

In a certain game of chance, a player has been observed to win five out of every nine games. In a particular tournament, three rounds are to be played. What are the outcomes and what are the probabilities for each outcome?

There are three stages and two outcomes at each stage - win, lose.

By reading along the branches of the tree, we see that there are eight outcomes:

win - win - win	win - win - lose
win - lose - win	win - lose - lose
lose - win - win	lose - win - lose
lose - lose - win	lose - lose - lose

The probabilities are multiplied along the branches because successive outcomes are independent. So, for example, we see that the probability of winning every game is $\frac{5}{9}\times\frac{5}{9}\times\frac{5}{9}=\frac{125}{729}$59​×59​×59​=125729​.

To find the probability of winning, say, two out of three games, we add the probabilities of each of the two-out-of-three outcomes. There are three of them, each with probability $\frac{100}{729}$100729​. So, the probability of winning two games out of three is $\frac{300}{729}=\frac{100}{243}$300729​=100243​ which is approximately $0.41$0.41.

Venn Diagrams

Another visual display that can help in probability questions is the Venn diagram. 

Subsets of outcomes with particular properties are displayed as areas within the diagram. Outcomes with two of the properties fall within the overlap of two areas. In the example above, the outcomes fall into the categories 'divisible by 2', 'divisible by 3' and 'divisible by neither 2 nor 3'.

Expectation

When a numerical result is associated with each outcome of an experiment, we can calculate a mathematical expectation.

Example 3

Suppose a team member in a sporting competition receives $currency(\$,800)$currency($,800) when the team wins and $currency(\$,500)$currency($,500) when the team loses. From a study of the relative frequencies of wins and losses over the previous few seasons, it is estimated that the team has a probability of $0.55$0.55 of winning any game. How much per game will the player receive on average in the long run?

We multiply the payment for winning by the probability of winning and also the payment for losing by the probability of losing. Thes are added to give the desired solution.

$800\times0.55+500\times0.45=665$800×0.55+500×0.45=665

The player will receive $currency(\$,665)$currency($,665) per game, on average.

More Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3

QUESTION 4