İSTATİSTİK I - Deneme Sınavı - 6

Consider any two experiments of which the first experiment can result in n1 and the second can result in n2 possible outcomes. Then, considering both experiments together, there are n1n2 possible outcomes. This principle can also be generalized to a finite number of experiments.

According to the basic principle of counting, it follows that there are in total 10.12=120 different possible classifications.

Let us denote the Italian models by X1 and X2, the Turkish models by Y1, Y2, and Y3. Then, there

E = { X1Y1, X1Y2, X1Y3, X2Y1, X2Y2, X2Y3}

There are different possible outcomes and the sample space is

S = {X1X2, Y1Y2, Y1Y3, Y2Y3, X1Y1, X1Y2, X1Y3, X2Y1, X2Y2, X2Y3}

If the event “one brand X and one brand Y phone is chosen” is denoted by E, the elements of E are .

E = { X1Y1, X1Y2, X1Y3, X2Y1, X2Y2, X2Y3}

The probability that event E will occur.

P(E)=n(E):n(S) = 6:10 =0.6

When the occurrence or non-occurrence of an event A does not affect the occurrence of another event B, then we say that A and B are statistically independent events.

An important application of conditional probability is given in the following result, which is known as Bayes’ Theorem.

Let E1, E2, ..., Ek be a collection of mutually exclusive and collectively exhaustive events. Then for any event A with P(A)≠0 and any i=1, 2..., k

P(Ei A)=P(Ei∩A)= P(AEi)P(Ei)
P (A) P (A E1 )P (E1 )+ P (A E2 )P (E2 )+!+ P (A Ek )P (Ek )

According to the Bayes's Theorem P(B|A) is equal to [P(A|B)*P(B)]/P(A). The values are given as so P(A|B)=0.30, P(B)=0.25 and P(A)=0.15. P(B|A)=(0.30*0.25)/(0.15) = 0.5. The answer is E.

According to the multiplication rule P(A∩B) = P(A⏐B)*P(B). The values are given as P(B)=0.45 and P(A⏐B)=0.20. P(A∩B) = (0.45)*(0.20) = 0.09. The answer is E.

For two dices the total number of possible outcomes is equal to 6*6 = 36. For the numbers' sum to be equal to 6 the number has to be (1,5), (2,4), (3,3), (4,2), (5,1) which shows that the number of times the event happening is equal to 5. The probability of two dices landing on numbers that when sum up is equal to 6 is 5/36. The answer is D.

A random experiment is any process that leads to two or more possible outcomes, without knowing exactly which outcome will occurIn a random experiment, Although we do not know which outcome will occur, we are able to list or describe all of the possible outcomes. The set of all possible outcomes is important to understand the random experiment. Therefore, it is called a sample space, which is restated below as a definition.

In the classical probability approach to assign a probability to an event, the assumption is that all the outcomes have the same chance of happening. Pick up a six-sided fair die, there are six numbers on each face of the die as 1, 2, 3, 4, 5, and 6. Classical probability says that each side of the die has the same chance to come face up if this die is thrown. Since there are 6 possible outcomes of throwing a six-sided fair die probability of obtaining any number represented on the faces of this six-sided fair die is 1/6.

In probability, if the occurrence of one event dictates that none of the other events can occur at the same time, we call this event mutually exclusive events.

In a study about college students, students are classified according to their department and preference of language course. If there are 10 different departments and 5 different language courses, in how many different ways can a student be classified? According to the basic principle of counting, it follows that there are in total 10.5=50 different possible classifications.

In a study about employess, employees are classified according to the department they work in and their preference of training courses. If there are 12 different departments and 7 different courses, according to the basic principle of counting, itf ollows that there are in total 12.7=84 different possible classifications.

When order is important, we call the arrangements of a finite number of distinct objects a permutation. For example, there are a total of 3!=6 different ordered arrangements of the three letters A,B, and C. If there are four letters (W,X,Y,Z) there are a total of 4!=24 different ordered arrangements.

This is a permutation with  repeated items. If all the seven letters in the given word were different, the total number of arrangements would be 6!. Since all the arrangements of the two letters C1 and C2 , and all the arrangements of the two letters M1 and M2 should be counted only once, it follows that the answer is: 6!/2!2!=180

We know that event B has occurred and we are interested in finding the probability of event A. That is, we are interested in finding the probability of A knowing that event B has occurred.The formula for this is as follows: P(A∩B) = P(A⏐B)P(B)

This is how we formulate the probability: P(A1∩ A2 ) = P(A2|A1 )P(A1 ) = P(A1 )P(A2|A1 ) 

This is how we calculate the probability of drawing two yellow cards: 6/36.5/35=1/42= 0.02381

Bayes’ Theorem is an important application of conditional probability. This theorem enables us to compute a particular conditional probability. Using Bayes’ Theorem, we can find the probability that a defective item is produced by a particular machine,

Suppose that we know that event B has occurred and we are interested in finding the probability of event A. That is, we are interested in finding the probability of A knowing that event B has occurred. The formula specified is actually called the multiplication rule and has important applications.

Discrete random variables have only a countable number of separate values such as 0, 1, 2 , 3... etc. For example, the number of students in a class for certain day or the number of customers in a supermarket after 5:00 PM are cases for discrete random variables since these variables are finite and countable. Conversely, continuous random variable can take entire infinite values in a given interval. Because of this reason, continuous random variables are commonly measured instead of counted. For instance, waiting time for customers in a supermarket cashier line and travel time of a bus between two points are examples for continuous random variables. The correct answer is D.