Supplementary Notes on probability, binomial distribution, testing of hypotheses

Question 1: There are 20 girls and 20 boys in class. The boys are asked to make a date with one girl. Assuming the choices are random and independent, what is the probability of a girl receiving no date?

Answer:

• The probability of receiving a date from any particular boy is 1/20 (i.e. 0.05). [The choices are random. ]
• The probability of not receiving a date from any particular boy is 0.95. [1 – 0.05]
• The probability of not receiving a single date from 20 boys will be (0.95)²⁰ [Each choice is independent of the other choices, i.e. multiplication rule for independent events], i.e. 0.36

Question 2: The probability of having 4 Aces in a draw of 5 cards.

Answer:

Method 1: Tree method:

Probability of having the first card not an Ace and the remaining cards as all Aces

+ probability of having the second card not an Ace and the other cards all Aces

+ probability of having the third card not an Ace and the other cards all Aces.

+ probability of having the fourth card not an Ace and the other cards all Aces

+ probability of having the first four cards as all Aces

= 48/52 × 4/51 × 3/50 × 2/49 × 1/48

+ 4/52 × 48/51 × 3/50 × 2/49 × 1/48

+ 4/52 × 3/51 × 48/50 × 2/49 × 1/48

+ 4/52 × 3/51 × 2/50 × 48/49 × 1/48

+ 4/52 × 3/51 × 2/50 × 1/49 × 48/48

= 5× 48× 4!/(52×51×50×49× 48) or 48×5!/(52× 51× 50× 49× 48) or 5!/(52× 51× 50× 49)

Method 2: Basic definition of probability

= number of possible events meeting the criteria
　　number of all possible events

= number of possible events with 4 aces in a draw of 5 cards
　　number of all possible events in a draw of 5 cards

= 　48 = 　　48 　　　　[To find the value of ⁵² C ₅ , see the attached notes on
　⁵² C ₅ 52!/[5!(52-5)!] 　　probability.]

= 48 × 5! × 47! = 5!/(52×51×50× 49)
　　52!

Note: the tree method is straight forward, but tedious. When the problem gets more complicated, this method is not practical. The method basing on the basic definition of probability is much simpler and powerful in solving this type of complicated problems.

Question 3:

There are 400 social work graduates looking for a job. There are 200 agencies having vacancies to recruit social workers and each agency has vacancies. Assuming that these agencies are recruiting at the same time, each social work graduate applies to 50 agencies and the distribution is random, each agency will interview only 10 applicants and the chances of getting an interview and getting a job offered are random.

1. What is the probability of a getting a job interview for each application?
2. How many job interviews will a graduate expect to get?
3. What is the probability of getting at least one job offer?

Answer:

1. The number of applications received by each agency = 400× 50/ 200 = 100
2. The probability of getting a job interview for each application = 10/100 = 0.1
3. The expected number of job interviews obtained by each graduate = 50 × 0.1 = 5
4. The probability of getting a job offer for each application = 0.1 × 0.2 = 0.02

The probability of not getting a job offer = 1- 0.02 = 0.98.

The probability of not getting a single job offer = 0.98⁵⁰ = 0.36.

The probability of getting at least one job offer = 1 – 0.36 = 0.64.

　

Question 4: Identify the number of possible routes from A to B, if you can only go left or down.

Answer:

Method 1: "Path Analysis"

1. Take the case of 1 × 1 first. From A to B, there are two possible routes. We mark the routes to B, "1" and "1", i.e. "2". The number of routes from A to B is 2.

2. Take the case of 2 × 2. The numbers marked on the routes represent the number of routes that can lead to that particular point. The number of routes leading to B will be 3+3 = 6.

3. Take the case of 3 × 3. The number of routes from A to B is 10+10 = 20.

4. Without much difficulty, you should be able to find in the case of a 4 × 4, the number of routes from A to B is 70.

5. The answer is 20 × 2× 70 = 2,800.

Note: The above method is only practical when n is small (where n represents the dimension of n × n). When n is very large, such as 100, the above method becomes very tedious. This is usual for this type of basic analysis. That is why the following two methods (inductive and deductive) are important. The basic analysis serves as the foundation to draw the conclusions of both the inductive and deductive analysis.

Method 2: Inductive (The path analysis forms the basis of inductive analysis)

From the analysis of method 1, we find the results follow the pattern of the Pascal's Triangle.

The number of paths in n×n table can be presented in the following table:

We can induce that for n ´ n table, the number of paths = ²ⁿC_n

Method 3: Deductive

For an n × n table, each path consists of n Horizontal segments and n Vertical segments. In other words, it is a combination of n H segments and n V segments.

Thus, the number of possible combinations is equal to ²ⁿC_n.

Take a special case, if it is a 3 × 4 table, the number of possible paths will be ⁷C₃ or ⁷C₄ (they are the same, i.e. 35). Take another special case, if the table is 2 × 5, the number of possible paths will be ⁷C₂ or ⁷C₅, i.e. 21.

　

Post-notes

Methods using step by step analysis as method 1 is always very useful for initial exploration.

The inductive method is always easier than the deductive method. In fact, it will be useful to start off with the inductive method, and after analyzing the problem, we may be able to use the deductive method. The deductive method is always more powerful.

The above methods are not only applicable to mathematical problems. They are also applicable to our daily life problems and our social work functions. For instance, a family seeks our assistance to solve its interpersonal problems. From this "case", we try to identify the concepts and possible theories underlying the interpersonal problems. After observing a number of cases, we begin to be able to draw commonalities (common characteristics) and begin to build theories to explain the phenomenon and guide our practice [Inductive Method]. Later on, we would be able derive to hypotheses from the theories we built and test them in real life situations [Deductive method].

　

Question 5：A couple had 6 children and all of them are girls. What would you say about the probability of having a boy or a girl if the couple decided to have another child?

Answer:

To begin the analysis, we can start with a hypothesis (i.e. usually called Null Hypothesis, usually denoted by H₀) that the probability of having a boy or a girl is fifty-fifty, i.e. 0.5.

Looking at the available information, i.e. the couple having 6 girls, the apparent alternative hypothesis is that the probability of having a girl for this couple is > 0.5. (In Statistics, this is called the Alternate Hypothesis, denoted by H₁ or H_A.)

Now, we know that this particular couple had 6 children and all are girls. Given if the probability of having a boy or a girl is 0.5, having all 6 girls, the probability will be equal to 1/2 ×1/2 ×1/2 ×1/2 ×1/2 ×1/2 [or (1/2 )⁶ ], i.e. 0.015625.

If the null hypothesis is correct, the probability of having 6 girls is so small (0.015625), we have sufficient confidence to reject the null hypothesis, and accept the alternate hypothesis, i.e. to conclude that the probability of having a girl for this couple is > 0.5.

We can also "bet" that it is more likely for this couple to have another baby girl if they attempt to have another child.

　

Other arguments

Superstition/Fatalistic: This couple is either very lucky or very unlucky (depending on how you look at it) for having all children being girls. Unless their "fate" change, their seventh baby will also be a girl. [Obviously, this argument is not scientific.]

Induction: Since this couple continuously bears 6 girls, it is likely to repeat for the seventh baby. [Inductive: in the sense that we predict what would happen in the future from what we observed in the past. We know that it is not a strong argument. What happened in the past may not happen in the future, and similarly what had not happened in the past may happen in the future.]

Independence: The probability of having a boy or girl for the seventh child is independent of the gender of the previous children, thus that chance of being a boy or a girl for the seventh child is still 0.5. [This is the confusion of the concept of "statistical independence" and "related". In Step 3 above, we have already applied the multiplication rule, i.e. assuming that each occurrence is statistically independent of the previous occurrence. Furthermore, we cannot say that the gender of the seventh child is in no-way related to the gender of the previous children. After all, they are children of the same parents. Insisting on that the probability is 0.5 is "stubborn" ignoring clear objective evidence of having 6 girls already.]

　

Notes on Statistics - Testing of Hypothesis

In Statistics, when the probability of observing a particular phenomenon under a null hypothesis is < 0.05, we say that the result is statistically significant, and we have sufficient confidence to reject the null hypothesis. The value 0.05 is sometimes called the significance level or critical value.

However, we should note that the conclusion is drawn as a matter of confidence not as a "proof". We have proved nothing. If the null hypothesis is true [probability = 0.5, or usually written as p = 0.5], we still have a chance of 0.015625 for having six girls in a series. Though the probability is small, it may still happen. This is called the Type I error, that is rejecting a null hypothesis when it is in fact true. In this case, to conclude that p > 0.5, we run the risk of 0.015625 in committing the Type I error.

If a particular couple has 4 children and are all girls, the probability under the null hypothesis of observing this is 1/2 × 1/2 × 1/2 × 1/2 [or (1/2 )⁴ ], i.e. 0.0625. Though 0.0625 is small, yet it is > 0.05, the significance level. Therefore, we do not reject the null hypothesis. By not rejecting the null hypothesis, we run another risk of error, that is not rejecting the null hypothesis when in act it is false. This is called Type II error.

　

Theoretically, we can never tell the probability of having a girl or a boy for a given couple, unless they bear infinite number of babies and then we can calculate the experimental probability. This is exactly the limit of our knowledge in science or social sciences. We are not able to obtain infinite number of observations and our scientific knowledge is always limited by the possible number of observations or experiments that we can have. Knowledge in pure mathematics is perhaps the only exception.