A random experiment is any process that leads to two or more possible outcomes, without knowing exactly which outcome will occur.

The set of all possible outcomes of a random experiment is called the sample space (S). Each possible outcome of a random experiment is called an elementary outcome(E).

If A and B are any two events then they are said to be mutually exclusive if AnB = Ø. Namely, the intersection set of events must be empty set if they are mutually exclusive.

Having a "head" and having a "tail" when throwing a coin are mutually exclusive.

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.

The empirical probability is based on experperiments.

Sometimes it may not possible to observe the outcomes of events; therefore, the researcher may assign a probability to an event. In subjective probability approach, the researcher assigns a suitable value as the probability of the event. Therefore, a personal judgement comes in to play to assign the probability.

When counting the possible outcomes of an event, it is sometimes important to distinguish between ordered and unordered arrangements. When order is important, we call the arrangements of a finite number of distinct objects a permutation. When order is not important, the arrangements of r objects from n distinct objects is called a combination.

Since the order of students is not important we can use combination. Thus:

C(10,3)=10!/7! 3!=7!*8*9*10/7!*3*2=4*3*10=120. So 120 different groups are possible.

In 10 days either 2 projects will finish(P(2)) or 1 project will finish(P(1)) or none (zero) project will finish (P(0)) in 10 days. Thus:

P(0)+P(1)+P(2)=1

We are interested in at least one project to finish. So we have to compute P(1)+P(2)

P(1)+P(2)=1-P(0)

Since P(Ali Finish)=0.4, then P(Ali Doesnt Finish)=1-0.4=0.6

Since P(Aysel Finish)=0.7, then P(Ali Doesnt Finish)=1-0.7=0.3

Thus P(0)=P(Neither Ali nor Aysel Finish)=06*0.3=0.18

Therefore:

P(1)+P(2)=1-P(0)=1-0.18=0.82

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. This is denoted by P(A|B) and is defined as:

P(A|B=P(AnB)/P(B)

In general, we say that the events A1, A2, ..., Ak are mutually statistically independent if and only ifP(A1 ?A2 ?... ?Ak) = P(A1)P(A2)...P(Ak)

If the dice is greater than 3, then the event space is

S=(4,5,6)

So, if we want it to be smaller than 6, we want the dice come 4 or 5. Therefore the probability is 2/3.

The probability of not being pink=13/20

The probability of being blue=5/20

Then probability of being blue, given not being pink=(5/20)/(13/20)=5/13

Since 1/4 of males have glasses, there are 1/4*20=5 males with glasses and 15 without glasses.

Since 1/3 of females have glasses, there are 1/3*30=10 females with glasses and 20 without glasses. Thus we can prepare the following table. From the table it is easily seen that 2/3 of students with glasses are female.(10/15=2/3)

Let P(A) denote having a head and P(B) having a number smaller than 3.

P(A)=1/2

P(B)=1/3 (since we want either 1 or 2 come on dice; 2/6=1/3)

Since these are independent events P(AnB)=P(A)*P(B)=1/2*1/3=1/6

P(Red/Model C)=15/50=0.3

P(Model C/Red)=15/40=0.375