### SAMPLE PROBLEMS ON THEORETICAL PROBABILITIES

Problems Set I: The Probability of a Single Event

Sample Problem #1
An experiment consists of tossing 4 coins simultaneously, once. Find the probability that at least two heads (H) appear.

The sample space is S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT}.
$\mid S \mid = 16$
The event is E = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH}
$\mid E \mid = 11$
$P(E) = \frac{\mid E \mid}{\mid S \mid} = \frac{11}{16} = 0.6875$
So, the probability that at least two heads appear is 0.6875.

Sample Problem #2
Fifteen cards are numbered from 1 to 15. The experiment consists of picking at random a card from the set of cards. Find the probability of getting a card with a number which is a multiply of 3.

The sample space is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
$\mid S \mid = 15$
The event is E  = {3, 6, 9, 12, 15}
$\mid E \mid = 5$
$P(E) = \frac{\mid E \mid}{\mid S \mid} = \frac{5}{15} = \frac{1}{3} \approx 0.3333$.
So, the probability of getting a card with a number which is a multiply of 3 is 0.3333.

Sample Problem #3
In the experiment of tossing two dice simultaneously once, find the probability that they show the numbers which differ by 4.

The sample space is S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}.
$\mid S \mid = 36$.
The event is E = {(1,5), (5,1), (2,6), (6,2)}
$\mid E \mid = 4$.
$P(E) = \frac{\mid E \mid}{\mid S \mid} = \frac{4}{36} = \frac{1}{9} \approx 0.1111$.
So, the probability that the dice show the numbers which differ by 4 is 0.1111.

Problems Set II: The Probability of Disjoint or Overlapping Events

Sample Problem #4
Two dice are rolled simultaneously once. Find the probability that the first dice show the side with 2 spots or the second dice with 5 spots.

The sample space is S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}.
$\mid S \mid = 36$.
Let A = the first dice show the side with 2 spots and B = the second dice show the side with 5 spots.
Then, A = {(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)}
B = {(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)}
$A \cap B = \{ (2,5) \}$.
Consequently,
$\mid A \mid = 6$ , $P(A) = \frac{6}{36} = \frac{1}{6}$
$\mid B \mid = 6$ , $P(B) = \frac{6}{36} = \frac{1}{6}$
$\mid A \cap B \mid = 1$, $P(A \cap B) = \frac{1}{36}$
Use the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, then we have:
$P(A \cup B) = \frac{1}{6} + \frac{1}{6} - \frac{1}{36} = \frac{11}{36} \approx 0.3056$
So,  the probability that the first dice show the side with 2 spots or the second dice with 5 spots is 0.3056.

Sample Problem #5
There are 10 balls in a box which are distinguishable only by the colours. Three of them are red, two are yellow, and the rests are green. One ball is drawn at random from the box. Calculate the probability that it is red or yellow.

From the text it can be inferred that the box contains 3, 2, and 5 red, yellow, and green balls, respectively. Name the balls R1, R2, R3, Y1, Y2, G1, G2, G3, G4, and G5 where R stands for red, Y yellow, and G green. The sample space is then S = {R1, R2, R3, Y1, Y2, G1, G2, G3, G4, G5}. Let A = a red ball is selected and B = a yellow ball is selected. Then A = {R1, R2, R3}, B = {Y1, Y2}, and $A \cap B = \{ \: \}$. From these we have $P(A) = \frac{3}{10}$, $P(B) = \frac{2}{10}$. As $A \cap B = \{ \: \}$, A and B are mutually exclusive (disjoint), so $P(A \cup B) = P(A) + P(B)$ applies. So, $P(A \cup B) = \frac{3}{10} + \frac{2}{10} = \frac{1}{2} = 0.5$. The probability of picking a red or yellow ball is 0.5.

Sample Problem #6
An experiment consists of tossing 4 coins simultaneously, once. Find the probability that at least two heads (H) appear or at least two tails (T) appear.

The sample space is S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT}.
$\mid S \mid = 16$
Let A = at least two heads appear and B = at least two tails appear
A = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH}
B = {HHTT, HTHT, THHT, HTTH, THTH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT}
$A \cap B = \{ HHTT, HTHT, THHT, HTTH, THTH, TTHH \}$
Then, $\mid A \mid = \mid B \mid = 11$ and $\mid A \cap B \mid = \{ HHTT, HTHT, THHT, HTTH, THTH, TTHH \}$.
$P(A) = P(B) = \frac{11}{16}$
$P(A \cap B) = \frac{6}{16}$
Applying the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, we have:
$P(A \cup B) = \frac{11}{16} + \frac{11}{16} - \frac{6}{16} = 1$
So, the probability that at least two heads (H) appear or at least two tails (T) appear is 1, i.e. the event will occur with certainty.

Problems Set III: The Probability of Independent or Dependent Events

Sample Problem #7
One bag contains 4 blue balls and 3 red balls, and a second bag contains 5 blue balls and 7 red balls. One ball is drawn at random from the first bag and is placed unseen in the second bag. What is the probability that the ball drawn from the first bag and the one from the second bag are both blue?

Let the balls in the first bag be B1, B2, B3, B4, R1, R2, R3 and the ones in the second be b1, b2, b3, b4, b5, r1, r2, r3, r4, r5, r6, r7.
Let A = a blue ball is drawn from the first bag and B = a blue ball is drawn from the second bag
The probability to find is $P(A \cap B)$, so we use the formula $P(A \cap B) = P(A) \cdot P(B \mid A)$.
To calculate P(A), we have to determine the sample space S1 and the event A.
S1 = {B1, B2, B3, B4, R1, R2, R3} and A = {B1, B2, B3}
Therefore, $\mid S_{1} \mid = 7$ , $\mid A \mid = 3$ , and $P(A) = \frac{3}{7}$.
To  calculate  $P(B \mid A)$, we have to assume that A occurs, i.e. a blue ball is drawn from the first bag. Based on the way how the experiment is conducted, the ball is placed in the second bag. So now there are 6 blue and 7 red balls, respectively, in the second bag. The sample space of the event $B \mid A$ is S2 = {b1, b2, b3, b4, b5, bx, r1, r2, r3, r4, r5, r6, r7}. In S2, bx represents the blue ball coming from the first bag; it is one of the four balls in the first bag: B1, B2, B3, and B4.
$\mid S_{2} \mid = 13$
$B \mid A = \{ b1, b2, b3, b4, b5, bx \}$
$\mid B \mid A \mid = 6$ , $P(B \mid A) = \frac{6}{13}$
Applying the formula $P(A \cap B) = P(A) \cdot P(B \mid A)$, we have:
$P(A \cap B) = \frac{3}{7} \cdot \frac{6}{13} = \frac{18}{91} \approx 0.1978$.
So, the probability that the ball drawn from the first bag and the the second are both blue is 0.1978.

Sample Problem #8
[Compare this problem with Problem #9.]
One bag contains 2, 5, and 3 red, blue, and green balls, respectively. Three balls are to be drawn from the bag in succession. Before the next ball is drawn, the previous ball selected is not put back into the bag. We call this sampling without replacement. What is the probability that the third ball we get is blue if the first ball is red and the second is green.

If the first ball we get is red and the second is green, then there are now 1, 5, and 2 red, blue, and green balls in the box, respectively. Name the ball R, B1, B2, B3, B4, B5, G1, G2, so the sample space of the third event (= getting a blue ball at the third drawing) is S = {R, B1, B2, B3, B4, B5, G1, G2}. The third event is E = {B1, B2, B3, B4, B5}. As $\mid S \mid = 8$ and $\mid E \mid = 5$, $P(E) = \frac{5}{8} = 0.625$. So the probability that the third ball we get is blue if the first ball is red and the second is green is 0.625.

Sample Problem #9
One bag contains 2, 5, and 3 red, blue, and green balls, respectively. Three balls are to be drawn from the bag in succession. The sampling is without replacement. What is the probability that we get red, green, and blue balls in succession?

Let A = a red ball is selected at the first drawing, B = a green ball is selected at the second drawing, and C = a blue ball is selected at the third drawing. The probability to find in this case is $P(A \cap B \cap C)$.
$P(A \cap B \cap C) = P(A \cap B) \cdot P(C \mid A \cap B)$ ........................................................ (*)
Since $P(A \cap B) = P(A) \cdot P(B \mid A)$, we can state (*) as:
$P(A \cap B \cap C) = P(A) \cdot P(B \mid A) \cdot P(C \mid A \cap B)$
Because initially there are 2, 5, and 3 red, blue, and green balls in the bag, respectively, the probability that a red ball is selected at the first drawing is $P(A) = \frac{2}{10} = \frac{1}{5}$. To calculate $P(B \mid A)$, we have to assume that A occurs. So now in the bag there are 1, 5, and 3 red, blue, and green balls, respectively. In this new condition, the probability of getting a green ball is $\frac{3}{9} = \frac{1}{3}$. So $P(B \mid A) = \frac{1}{3}$. To calculate $P(C \mid A \cap B)$, we have to assume that $A \cap B$ occurs, i.e. we get a red ball at the first drawing and a green one at the second. So now in the bag there are 1, 5, and 2 red, blue, and green balls, respectively. In this new condition, the probability of getting a blue ball is $\frac{5}{8}$. So $P(C \mid A \cap B) = \frac{5}{8}$. Substituting suitable values for P(A), $P(B \mid A)$, and $P(C \mid A \cap B)$ in (*), we get:
$P(A \cap B \cap C) = \frac{1}{5} \cdot \frac{1}{3} \cdot \frac{5}{8} = \frac{1}{24} \approx 0.0417$.
So, the probability that we get red, green, and blue balls in succession is 0.0417.

### ANATOMY OF FREQUENCY DISTRIBUTION TABLES

Quantitative data can be presented in the form of frequency distribution tables. Here is a typical frequency distribution table (FDT). Table 1 shows  the duration of 20 conversations. There are five classes (or categories) for the duration. Each class is assigned to a symbol called class interval . The class interval of the first class is 5 - 9 minutes, the class interval of the second class is 10 - 14 minutes, etc. The frequencies in the table indicate how many conversations fall in each category/class. For example, there are 3 conversations whose duration is between 5 and 9 minutes, 5 conversations whose duration is between 10 and 14 minutes, etc. Class Limits There are five classes in Table 1. Each class has a lower class limit (LCL) and an upper class limit (UCL). LCL of class #1 is 5 minutes and the UCL is 9 minutes. LCL of class #2 is 10 minutes and the UCL is 14 minutes, etc.  Class Boundaries Each class has a lower class boundary (LCB) and an upper class boundary

### CONSTRUCTING THE FREQUENCY DISTRIBUTION TABLE

The frequency distribution table is a table that divides data into groups (classes) and shows how many data values occur in each group/class. Below is an example of frequency distribution table. Now we are learning how to create a frequency distribution table. Suppose we have a collection of ungrouped data on last year’s advertising expenditures of 40 logistics companies, recorded in millions Rupiahs. To construct a frequency distribution table of the ungrouped data, apply the following steps. Step 1: Find the range of the data The range (R) is defined as the difference between the largest data and the smallest data. In this case, R = 307 - 242 = 65. Step 2: Determine the number of categories/classes (k) Applying Sturges rule (k = 1 + 3,322 log n, where n = the number of data), we have: $k = 1 + 3.322 \: log \: 40 \approx 6.32$ As the value of k must be a natural number, 6.32 is rounded up to 7, so k = 7. Step 3: Determine the class width (c) To find c, use \$

### BINOMIAL DISTRIBUTION

Suppose that a fair coin is tossed 20 times. The word ‘fair’ here means that each side of the coin has the equal probability of appearing. If it turned out that the Head (H) side appeared 18 times out of the 20 times of tossing, we would possibly be surprised. Why? Although we  know that it can happen by chance, it happens very rarely. It is unlikely, but not totally impossible. If the coin is fair indeed, we expect that the frequency of occurrence of each side is around 10. Statistically speaking, the probability that the H side appears 18 times is very small. Furthermore, the probability of H side appearing 18 times is much smaller than 10 times appearing. If the coin is fair, it is more expectable that the H side appears 10 times than it does 18 times. Suppose that out of the 20 times tossing, we get 13 times occurrence of H side. Still we may not be so surprised by the result as we may if the H side appears 18 times. However, if the coin is fair the probability that the H side ap