5.3.1 Evaluating Probabilities in Simple Cases

In this topic we will learn how to:

• evaluate probabilities in simple cases by means of enumeration of equiprobable elementary events, or by calculation using permutations or combinations
• use addition and multiplication of probabilities, as appropriate, in simple cases

In this topic we will delve into the basics of AS level probability. Let’s learn how to deal with probabilities in simple instances by walking through past paper questions.

1. Three coins A, B and C are each thrown once. Coins A and B are each biased so that the probability of obtaining a head is $\frac{2}{3}$. Coin C is biased so that the probability of obtaining a head is $\frac{4}{5}$. Show that probability of obtaining exactly $2$ heads and $1$ tail is $\frac{4}{9}$. (9709/53/O/N/20 number 6)

We have to consider the following scenarios,
Coin A shows head, Coin B shows head and Coin C shows tail
Coin A shows head, Coin B shows tail and Coin C shows head
Coin A shows tail, Coin B shows head and Coin C shows head

Let’s start by considering the first scenario, Coin A shows head, Coin B shows head and Coin C shows tail,
$$P(HHT) = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{5}$$Note: Since the probability of Coin C showing head is $\frac{4}{5}$ then its probability of showing a tail is $\frac{1}{5}$.

Which simplifies to give,
$$P(HHT) = \textcolor{#2192ff}{\frac{4}{45}}$$Let’s consider the second scenario, Coin A shows head, Coin B shows tail and Coin C shows head,
$$P(HTH) = \frac{2}{3} \times \frac{1}{3} \times \frac{4}{5}$$Note: Since the probability of Coin B showing head is $\frac{2}{3}$ then its probability of showing a tail is $\frac{1}{3}$.

Which simplifies to give,
$$P(HTH) = \textcolor{#0f0}{\frac{8}{45}}$$Let’s consider the third scenario, Coin A shows tail, Coin B shows head and Coin C shows head,
$$P(THH) = \frac{1}{3} \times \frac{2}{3} \times \frac{4}{5}$$Note: Since the probability of Coin A showing head is $\frac{2}{3}$ then its probability of showing a tail is $\frac{1}{3}$.

Which simplifies to give,
$$P(HTH) = \textcolor{red}{\frac{8}{45}}$$Putting all three scenarios together,
$$P(2H \textmd{ and } 1T) = \textcolor{#2192ff}{\frac{4}{45}} + \textcolor{#0f0}{\frac{8}{45}} + \textcolor{red}{\frac{8}{45}}$$$$P(2H \textmd{ and } 1T) = \frac{4}{9}$$Therefore, the final answer is,
$$\frac{4}{9}$$2. Out of a class of $8$ boys and $4$ girls, a group of $7$ people is chosen at random. Find the probability that the group of $7$ includes one particular boy.(9709/63/O/N/18 number 4)

This is an example of a probability that involves combinations. To find the probability that the group includes one particular boy, we have to consider the total number of selections possible and then all the selections that include the particular boy,
$$P(\textmd{group with particular boy}) = \frac{\textmd{number of selections with particular boy}}{\textmd{total number of selections possible}}$$Let’s assign the particular boy a letter of $P$. Selections that include $P$ would look something like this,
$$7 \textmd{ boys } \ \ \ \ 4 \textmd{ girls}$$$$11 \textmd{ people }$$$$\underline{P}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}$$Note: We have $11$ people because $P$ is removed from the pool of selection since we have already picked him.

That means we have $6$ slots for $11$ people,
$$^{11}C_{6}$$Which simplifies to give,
$$\textcolor{#2192ff}{462}$$Now let’s find the total number of selections possible for $7$ people from a total of $12$ people,
$$^{12}C_{7}$$Which simplifies to give,
$$\textcolor{#0f0}{792}$$Putting everything together,
$$P(\textmd{group with particular boy}) = \frac{\textmd{number of selections with particular boy}}{\textmd{total number of selections possible}}$$$$P(\textmd{group with particular boy}) = \frac{\textcolor{#2192ff}{462}}{\textcolor{#0f0}{792}}$$$$P(\textmd{group with particular boy}) = \frac{7}{12}$$Therefore, the final answer is,
$$\frac{7}{12}$$3. Two ordinary fair dice, one red and the other blue, are thrown. Event A is the ‘score on the red die is divisible by $3$‘. Event B is the ‘the sum of the two scores is at least $9$‘. Find $P(A \cap B)$. (9709/51/O/N/20 number 1)

To be able to solve this question we have to draw a possibility space diagram. It is a diagram that shows all the possible outcomes for particular events. In our case, for events A and B. To draw the possibility space diagram, label the $x$-axis as red die and the $y$-axis as blue die. On the respective axes label the possible scores on a fair die i.e $1 - 6$. Plot a dot for every possible outcome for the two dice. Identify and mark events A and B on the possibility space diagram. Insert a key to enable the reader to understand what particular labels mean.

The below is an example of a possibility space diagram relevant to this question,

From the possibility space diagram, we can deduce that,
$$P(A \cap B) = \frac{5}{36}$$Therefore, the final answer is,
$$\frac{5}{36}$$