Thursday, October 22, 2015

Equiprobable events (equally probable events).

You toss an unbiased coin 5 times. Is HHHHH (5 heads in a row) as probable as HTHTH?Answer: Yes.
On the first throw H or T is equally likely. Say it landed H. The coin does not remember what it did on the first throw, so for the next throw H or T is equally likely, so HH is as likely as HT. One can go on in this manner to show that HHHHH is as likely as HTHTH. The probability of HHHHH is 0.5x0.5x0.5x0.5x0.5 This is the same probability as the probability for HTHTH. Equiprobable events are very important in statistics as many problems are solved using equiprobable events.

Mutually Exclusive versus Independent

Say you have 3 different types of equal size marbles in a sack and there are equal numbers of each marble. Call them marbles A, B and C. Now you draw a marble out, look at it and replace it What is the probability of drawing out a marble of type A? Answer 1/3.
What is the probability of drawing out a marble of type A or B? Answer 2/3. Now this is important. We have a SINGLE OUTCOME (draw a marble) and we are considering the probabilities of it being of type A or B (the outcome A and B are mutually exclusive because they cannot both happen with our SINGLE OUTCOME). We add 1/3 to 1/3 and get 2/3 for the probability of drawing type A or B. This is the addition rule for mutually exclusive events.
Now we ask another type of question. If we draw a marble and consider this to be an event (and then replace it) and then draw another marble (and replace it) and consider that to be another event we can ask ourselves what the probability is that we will draw a type B and then a type C.. The TWO events are independent and the answer is (1/3)(1/3) =1/9. The events are independent because the probability remains 1/3 (the probability on the second draw is not affected by the first draw).
One point here: The mutually exclusive idea applies to a single outcome. The independence idea refers to TWO different outcomes each from a different event. When you add probabilities you are considering the outcomes that could occur where there is only ONE outcome (it could be type A or Type B or Type C, but only one of them occurs).
Now consider this example. When you drop an egg from a low height there is a probability of 0.6 that the egg will break. You drop 4 eggs and view the result (outcome). Again there is only one outcome, but various possibilities for that outcome. OUTCOME FOR 2 Eggs breaking: One outcome could be that 2 eggs break in this way: 1 breaks 2 does not, 3 breaks, 4 does not OR another outcome could be 1 does not, 2 does not 3 does and 4 does. You have a mutually exclusive situation because the outcomes are different, BUT THERE IS ONLY ONE OUTCOME. The outcome 1 breaks and 2 breaks, but the others do not is different to the outcome that 1 does not, 2 does not and 3 and 4 do (they are mutually exclusive because for your ONE outcome only one of them can occur). You can find the probability of two eggs breaking by using the MUTUALLY EXCLUSIVE RULE and adding the probability that 1 and 2 break, others do not to 1 and 3 break and others do not and so on until you have added all probabilities for cases where two eggs could break. To summarize: Where there is one outcome, but various mutually exclusive ways that single outcome could occur, you add the mutually exclusive probabilities. Now consider the independent events associated: Example: What is the probability that egg 1 breaks and egg 2 does not and egg 3 does and egg 4 does not. This applies to different events and the answer is (0.6)(0.4)(0.6)(0.4) using the multiplication rule for different independent events. The events are independent because the breaking of one egg does not affect the probability that another egg will break.

To summarise: One outcome, but various possibilities that are mutually exclusive – use the addition rule. Two or more independent events (one outcome for each of the events) use the multiplication rule.

Saturday, January 16, 2010

Daily mathematics or physics hint.

For mathematics and physics help see Mathematics and physics help on Fiverr I have a BSc from UNISA (graduated 1982).
2) Daily hint 27 June 2015. A 70 kg man is standing still on a rough floor. What is the resultant force on the man? Answer: Because the man is not accelerating we know that the sum of all forces (resultant force) acting on the man is zero. There is the gravitational force downwards and the force of the floor upwards. Their vector sum is zero. Note that F=ma says that the sum of all forces acting on a body equals the mass of the body times the acceleration of that body.
1) Daily hint 26 June 2015:
F = (kq1q2)/r
Above is the mathematical statement. In words: The force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The constant k is always the same from problem to problem. It is called Coulomb's constant or called the electric force constant or electrostatic constant.
If you had to make r the subject of the formula how would you go about it? You could multiply both sides by r2 and then divide both sides by F and then take the square root of both sides (or multiply both sides by r2/F and take the square root of both sides).