# 0.11 Independent and dependent events  (Page 2/2)

 Page 2 / 2

## Use of a venn diagram

We can also use Venn diagrams to check whether events are dependent or independent.

Independent events

Events are said to be independent if the result or outcome of one event does not affect the result or outcome of the other event. So P(A/C)=P(A), where P(A/C) represents the probability of event A after event C has occured.

Dependent events

Two events are dependent if the outcome of one event is affected by the outcome of the other event i.e. $P\left(A/C\right)\ne P\left(A\right)$

. Also note that $P\left(A/C\right)=\frac{P\left(A\cap C\right)}{P\left(C\right)}$ . For example, we can draw a Venn diagram and a contingency table to illustrate and analyse the following example.

A school decided that its uniform needed upgrading. The colours on offer were beige or blue or beige and blue. 40% of the school wanted beige, 55% wanted blue and 15% said a combination would be fine. Are the two events independent?

1.  Beige Not Beige Totals Blue 0,15 0,4 0,55 Not Blue 0,25 0,2 0,35 Totals 0,40 0,6 1
2. P(Blue)=0,4, P(Beige)=0,55, P(Both)=0,15, P(Neither)=0,20

Probability of choosing beige after blue is:

$\begin{array}{ccc}\hfill P\left(Beige/Blue\right)& =& \frac{P\left(Beige\cap Blue\right)}{P\left(Blue\right)}\hfill \\ & =& \frac{0,15}{0,55}\hfill \\ & =& 0,27\hfill \end{array}$
3. Since $P\left(Beige/Blue\right)\ne P\left(Beige\right)$ the events are statistically dependent.

## Applications of probability theory

Two major applications of probability theory in everyday life are in risk assessment and in trade on commodity markets. Governments typically apply probability methods in environmental regulation where it is called “pathway analysis”, and are often measuring well-being using methods that are stochastic in nature, and choosing projects to undertake based on statistical analyses of their probable effect on the population as a whole. It is not correct to say that statistics are involved in the modelling itself, as typically the assessments of risk are one-time and thus require more fundamental probability models, e.g. “the probability of another 9/11”. A law of small numbers tends to apply to all such choices and perception of the effect of such choices, which makes probability measures a political matter.

A good example is the effect of the perceived probability of any widespread Middle East conflict on oil prices - which have ripple effects in the economy as a whole. An assessment by a commodity trade that a war is more likely vs. less likely sends prices up or down, and signals other traders of that opinion. Accordingly, the probabilities are not assessed independently nor necessarily very rationally. The theory of behavioral finance emerged to describe the effect of such groupthink on pricing, on policy, and on peace and conflict.

It can reasonably be said that the discovery of rigorous methods to assess and combine probability assessments has had a profound effect on modern society. A good example is the application of game theory, itself based strictly on probability, to the Cold War and the mutual assured destruction doctrine. Accordingly, it may be of some importance to most citizens to understand how odds and probability assessments are made, and how they contribute to reputations and to decisions, especially in a democracy.

Another significant application of probability theory in everyday life is reliability. Many consumer products, such as automobiles and consumer electronics, utilize reliability theory in the design of the product in order to reduce the probability of failure. The probability of failure is also closely associated with the product's warranty.

## End of chapter exercises

1. In each of the following contingency tables give the expected numbers for the events to be perfectly independent and decide if the events are independent or dependent.
1.  Brown eyes Not Brown eyes Totals Black hair 50 30 80 Red hair 70 80 150 Totals 120 110 230
2.  Point A Point B Totals Busses left late 15 40 55 Busses left on time 25 20 45 Totals 40 60 100
3.  Durban Bloemfontein Totals Liked living there 130 30 160 Did not like living there 140 200 340 Totals 270 230 500
4.  Multivitamin A Multivitamin B Totals Improvement in health 400 300 700 No improvement in health 140 120 260 Totals 540 420 960
2. A study was undertaken to see how many people in Port Elizabeth owned either a Volkswagen or a Toyota. 3% owned both, 25% owned a Toyota and 60% owned a Volkswagen. Draw a contingency table to show all events and decide if car ownership is independent.
3. Jane invested in the stock market. The probability that she will not lose all her money is 0,32. What is the probability that she will lose all her money? Explain.
4. If D and F are mutually exclusive events, with P(D ${}^{\text{'}}$ )=0,3 and P(D or F)=0,94, find P(F).
5. A car sales person has pink, lime-green and purple models of car A and purple, orange and multicolour models of car B. One dark night a thief steals a car.
1. What is the experiment and sample space?
2. Draw a Venn diagram to show this.
3. What is the probability of stealing either a model of A or a model of B?
4. What is the probability of stealing both a model of A and a model of B?
6. The probability of Event X is 0,43 and the probability of Event Y is 0,24. The probability of both occuring together is 0,10. What is the probability that X or Y will occur (this inculdes X and Y occuring simultaneously)?
7. P(H)=0,62, P(J)=0,39 and P(H and J)=0,31. Calculate:
1. P(H ${}^{\text{'}}$ )
2. P(H or J)
3. P(H ${}^{\text{'}}$ or J ${}^{\text{'}}$ )
4. P(H ${}^{\text{'}}$ or J)
5. P(H ${}^{\text{'}}$ and J ${}^{\text{'}}$ )
8. The last ten letters of the alphabet were placed in a hat and people were asked to pick one of them. Event D is picking a vowel, Event E is picking a consonant and Event F is picking the last four letters. Calculate the following probabilities:
1. P(F ${}^{\text{'}}$ )
2. P(F or D)
3. P(neither E nor F)
4. P(D and E)
5. P(E and F)
6. P(E and D ${}^{\text{'}}$ )
9. At Dawnview High there are 400 Grade 12's. 270 do Computer Science, 300 do English and 50 do Typing. All those doing Computer Science do English, 20 take Computer Science and Typing and 35 take English and Typing. Using a Venn diagram calculate the probability that a pupil drawn at random will take:
1. English, but not Typing or Computer Science
2. English but not Typing
3. English and Typing but not Computer Science
4. English or Typing

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