PROBABILITY

This topic is all about gaining a core understanding of the concept. There are no complex formulae to memorize, it is just how well you grasp the procedures. We will help you build an intuition by not just saying what to do, but explaining why we do it.

Introduction

Probability is simply how likely something is to happen.

The probability of an event A is often written as P(A).

There is one main formula that will be used throughout this topic: $P(\text{event})=\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$ This forms the basis for all probability questions, it just takes a little bit of work to get there in some cases.

The best example for understanding probability is flipping a coin:

There are 2 possible outcomes— heads and tails.

What is the probability of the coin landing on heads? You might intuitively know that the likelihood of landing on one of the sides is half/half, or 50%. This is calculated as $P(Heads)=\displaystyle\frac{1}{2}$ . There is one favourable outcome (heads) and two possible outcomes (heads and tails).

Let us look at another example, a fair die. The word fair means it is equally likely to fall on any side. The probability of getting a 5 is: $P(5)=\displaystyle\frac{1}{6}$ , as there are 6 sides and 1 favourable outcome. Likewise, the probability of getting a 2 or 4 is $\displaystyle\frac{2}{6}$ .

If all the possible outcomes are favourable (i.e. the numerator equals the denominator), the probability is 1. Therefore, the probability of something that is guaranteed to happen is 1. But what if we look at the probability of an event A either happening or not happening? This makes all the outcomes favourable, since the two possible outcomes (A occuring or not occuring) are both favourable. Thus, the probability of A occuring or A not occuring is 1. From this, we can make an interesting observation: $P(A)+P(not\>A)=1$ $P(not\>A)=1-P(A)$ The probability of an event A not happening is $1-P(A)$ . For example, if the probability of a bus being late is 0.2, then the probability of it not being late is: $P(\text{not late})=1-0.2=0.8$

Example 1

A box contains 6 red balls and some blue ones. If a ball is picked at random the probability that it is red is 0.25. Find the number of blue balls.

Let the total number of balls be x. $P(red)= \frac{6}{x}$ $0.25=\frac{6}{x}$ $x=24$ The total number of balls is 24. Thus the total number of blue balls is $24-6=18$ .

Example 2

If a point is selected at random inside the bigger circle, what is the probability that it lies outside the smaller circle?

In this situation, we must consider the areas of the circles. The area of the larger circle is $25\pi$ , while that of the smaller circle is $5\pi$ . The probability of the point being in the smaller circle is: $\begin{split} \frac{\text{Area of smaller circle}}{\text{Area of larger circle}} & = \frac{5\pi}{25\pi} \\ & = \frac{1}{5}\end{split}$

Let’s take a short moment to find out how probability came to be. The earliest mathematical records of it stemmed from an interesting question, known as the problem of points. The problem concerns a game of chance with two players who have equal chances of winning each round. The players contribute equally to a prize pot, and agree in advance that the first player to have won a certain number of rounds will collect the entire prize. Now suppose that the game is interrupted before either player has achieved victory. How does one then divide the pot fairly?

Pierre de Fermat

This problem was posed to Blaise Pascal and Pierre de Fermat in 1654. They are credited as the fathers of probability. The two mathematicians figured out that what is important is not the number of rounds each player has won so far, but the number of rounds each player still needs to win in order to achieve overall victory. That is, the probability of winning. For example, a player with an 8-6 lead in a game to 10 has the same chance of winning as a player with an 18-16 lead in a game to 20. They used this reasoning to find the exact proportion in which the pot should be divided. A simplified version of this game is shown in the videos below, but feel free to explore this further on your own to see the exact solution they came up with.

Blaise Pascal

Combined Events

Combined events refer to various ways events can be related. We will discuss two types of combined events: mutually exclusive events and independent events.

Mutually Exclusive Events

This is a situation where 2 events cannot occur at the same time. If one event occurs, the other cannot.

A good example is a coin flip: if heads occurs, tails cannot also occur at the same time.

If A and B are mutually exclusive events: $P(A\,\,or\,B) = P(A) + P(B)$

For example, the probability of rolling a 3 or an even number on a fair die is: $\begin{align*}P(\text{3 or even})&=P(3)+P(even)\\&=\frac{1}{6}+\frac{3}{6}\\&=\frac{4}{6}\end{align*}$

Independent Events

2 events are independent if the occurrence of one does not influence the occurrence of the other.

For example, if I toss a coin two times, the outcome of the first toss (heads or tails) does not affect the outcome of the second toss. Similarly, the outcome of the second toss does not affect the first toss in any manner.

If A and B are 2 independent events, the probability of them both occurring is the product of their individual probabilities. $P (A \,\,and\,\, B) = P(A) \times P(B)$

Probability With Replacement and Without Replacement

This usually involves picking objects from a bag.

With replacement: Each time an object is removed, it is put back into the bag before the next time an object is removed again. The events are independent (the results of the first pick will not change the probabilities of the second pick).

Without replacement: Each time an object is taken out, it stays out as the next object is removed. The events are dependent (the result of the first pick affects the probabilities of the second pick).

In a situation with replacement, the denominator stays the same while calculating probability. In one without replacement, the denominator changes when calculating probability.

Example

A bag contains 8 black balls and 5 red ones. If 2 balls are drawn from the bag one at a time, find the probability of drawing balls of different colours:

a) without replacement.

There are 2 ways of getting balls of different colours. First, drawing a black then a red, or drawing a red then a black.

$\begin{split} P(B\ then\ R) & = \frac{8}{13} \times \frac{5}{12} \\& = \frac{40}{156}\end{split}$ $\begin{split} P(R\ then\ B) & = \frac{5}{13} \times \frac{8}{12} \\& = \frac{40}{156}\end{split}$
Since the 2 events are mutually exclusive, we add the probabilities to get: $\frac{40}{156}+\frac{40}{156}=\frac{80}{156}$

b) with replacement.

Since the balls are being replaced, the number of balls in the bag is always 13 when a ball is picked. Therefore: $\begin{split} P(B\ then\ R) & = \frac{8}{13} \times \frac{5}{13} \\& = \frac{40}{169}\end{split}$ $\begin{split} P(R\ then\ B) & = \frac{5}{13} \times \frac{8}{13} \\& = \frac{40}{169}\end{split}$ We add these probabilities to get: $\frac{40}{169}+\frac{40}{169}=\frac{80}{169}$

Tree Diagrams

Sometimes it is difficult to tell whether to multiply or add probabilities. A probability tree makes it easier to figure out when to add and when to multiply. Plus, seeing a diagram of your problem instead of a bunch of equations and numbers on a paper can help you see the problem more clearly.

A probability tree has two main parts: the branches and the ends. The probability of each branch is generally written on the branches, while the outcome is written on the ends of the branches.

Probability trees make the question of whether to multiply or add probabilities simple: multiply along the branches and add probabilities down the columns.

Example

If you toss a coin three times, what is the probability of getting:

(a) 3 heads?

We can draw a probability tree to solve this as follows:

i. Draw the branches for the first toss and indicate the probabilities on each branch.

ii. Draw the branches for the second toss and indicate the probabilities on each branch.

iii. Draw the branches for the final toss and indicate the probabilities on each branch.

iv. Write the outcomes at the ends of the final branches.

Now that we have the probability tree above, it is easy to determine how 3 heads can be obtained. The only way to get 3 heads is to follow the top branch every time, ending up with the outcome HHH. We multiply the probabilities across the branches to get the probability of getting 3 heads. $P(HHH)=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}$

(b) 2 tails and a head, in any order?

Using the probability tree, we find that HTT, THT and TTH all have 2 tails and a heads. As calculated in (a), the probability of each outcome is $\displaystyle\frac{1}{8}$ . Since the different outcomes are in a column, we add them. $P(HTT\ or\ THT\ or\ TTH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

Example KCSE Question

The table below shows the number of students in each class in a school. The percentage(%) of the students in each class who wear glasses is also shown.

Class	Form 1	Form 2	Form 3	Form 4
Number of students	60	56	44	40
Percentage (%) with glasses	10%	12.5%	25%	17.5%

a) A student is chosen at random from the school. Determine the probability that the student wears glasses.

The total number of students is $60+56+44+40=200$ . Out of these, the number that wears glasses is: $\frac{10}{100}\times60+\frac{12.5}{100}\times56+\frac{25}{100}\times44+\frac{17.5}{100}\times40=31$

Therefore, the probability that a student wears glasses is: $P(\text{glasses})=\frac{31}{200}$

(b) 2 students are chosen at random from the school. Determine the probability that one of the students is in Form 1 while the other is in Form 4 and both wear glasses.

This question is slightly involving, so pay close attention. Let us look at the first part of the question, picking a form 1 and a form 4 student.
The probability of picking a form 1 student is: $P(\text{Form 1})=\frac{60}{200}$
After choosing a form 1, there are 199 remaining students. Thus, when finding the probability that the next student is a form 4, the denominator is 199. $P(\text{Form 4})=\frac{40}{199}$

Now that we know this, we find the probabilities of each one having glasses. $P(\text{Form 1 with glasses})=\frac{60}{200}\times\frac{10}{100}=\frac{6}{200}$ $P(\text{Form 4 with glasses})=\frac{40}{199}\times\frac{17.5}{100}=\frac{7}{199}$

Now, we combine the 2 to get: $P(\text{F1g then F4g})=\frac{6}{200}\times\frac{7}{199}=\frac{21}{19900}$

Notice that the above only accounts for if a form 1 with glasses is chosen first, then a form 4 with glasses is chosen. However, we still have not considered the case where the form 4 is chosen, then the form 1. To do this: $P(\text{F4g then F1g})=\frac{7}{200}\times\frac{6}{199}=\frac{21}{19900}$

We then add these 2 cases to get: $\begin{split}P(\text{F1g and F4g or F4g and F1g}) &= \frac{21}{19900}+\frac{21}{19900} \\[10 pt] &\boxed{=\frac{21}{9950}}\end{split}$

But probability is not just limited to math textbooks. It has a variety of applications in the real world. An interesting application, the Monty Hall problem, is covered in the following video. It is a famous probability puzzle.