Probability

1.1 Probability

Probability is a branch of mathematics, which deals with the likelihood of events

happening. An Event is often a set of outcomes of an experiment. Probability
works because of the nature of data, which can be discrete or continuous and thus
can be quantified. An example of an experiment could be tossing of a coin 20
times, and recording the results of it and tabulating them. The list of outcomes
becomes the event in probability.

1.2 Events

The events that we look at, at Matric level are Inclusive, Mutually Exclusive
Events, Exhaustive, Complementary Events, Dependent Events and
Independent Events.

1.3 Examples of Events and Applications

1.3.1 Inclusive Events

These events can occur at the same time. The following rule applied when dealing
with inclusive events:

𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 𝑎𝑛𝑑 𝐵)

We have to subtract the probability of A and B since it will be counted twice in

adding the two probabilities. You know an event is inclusive when 𝑃(𝐴 𝑜𝑟 𝐵) > 0

Alternative notation using Venn diagram notation is:

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)

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 A B
1 4
2 3 5
6 7

1.3.2 Mutually Exclusive Events

Unlike Inclusive events, Mutually Exclusive events are those that cannot occur at
the same time. Therefore, the intersection of these events is zero. Using Probability
Notation, it follows:

𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵)

A B
1 4
1.3.3 Dependent Events
2 5

1.3.4 Independent Events 3

For this case: 𝑃(𝐴 ∩ 𝐵) = 0.

1.3.3 Exhaustive Events

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In a given sample space (Universal set) for exhaustive events it is such that:
𝑃(𝐴 𝑜𝑟 𝐵) = 1. The Venn diagram for it looks like this:

A B
1 4
3
2 5

For Exhaustive events there are no elements outside of space A or B as with

previous other cases.

1.3.4 Complementary Events

Mutually Exclusive, Exhaustive events are called complementary events such that:
𝑃(𝑛𝑜𝑡 𝐴) 𝑤ℎ𝑖𝑐ℎ 𝑐𝑎𝑛 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠 𝑃(𝐴′ ) is the compliment of A.

𝑃(𝑛𝑜𝑡𝐴) + 𝑃(𝐴) = 1

From the above it can be concluded that 𝑃(𝑛𝑜𝑡 𝐴) = 1 − 𝑃(𝐴)

Using the Venn diagram notation, it can be seen that:

A B
1 4
2 5
3 7
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Since 𝑃(𝐴) + 𝑃(𝐵) = 1 𝑖𝑡 𝑚𝑒𝑎𝑛𝑠 𝐴 𝑎𝑛𝑑 𝐵 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦

1.4 Tree Diagrams

There are some experiments that have to be done repeatedly then we can determine
the probability of an even. A typical example is the tossing of a coin three times.
There are eight possible results that can be gotten from such an experiment and
tree diagrams make it easier to identify the probability space. An example of a tree
diagram being used is below:

1.4.1 Example 1: Tree Diagram

A coin was tossed two times; represent this information on a tree diagram

Using the results of the sample space is S = {HH, HT, TH, TT}

Therefore, we can use the results from the sample space to calculate the different
probabilities. In this case the probability of getting a head or a tail is:

1 1 2 1
𝑃(𝐻 𝑜𝑟 𝑇) = 𝑃(𝐻𝑇) + 𝑃(𝑇𝐻) = + = =
4 4 4 2
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1.5 Dependent and Independent Events

Tree diagrams are useful when dealing with Independent events such as the tossing
of a coin and rolling a dice. These two events are independent since they can
happen simultaneously since the outcome of the first even has no effect on the
outcome of the second event.

On the other end, dependent events are such that the occurrence of one event
depends on what has happened with another event. An example of dependent
events is to draw two diamond cards consecutively from a deck of 52 cards
without replacement. Since the probability space reduces from the first event
(picking a card) it follows these are dependent events.

1.5.1 Example of Independent Events

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𝑬𝒙𝒂𝒎𝒑𝒍𝒆 𝟏. 𝟏 𝑎 𝐶𝑜𝑖𝑛 𝑤𝑎𝑠 𝑡𝑜𝑠𝑠𝑒𝑑 𝑎𝑛𝑑 𝑎 𝑑𝑖𝑒 𝑤𝑒𝑟𝑒 𝑟𝑜𝑙𝑙𝑒𝑑 𝑜𝑛𝑒 𝑎𝑓𝑡𝑒𝑟 𝑎𝑛𝑜𝑡ℎ𝑒𝑟
𝑎𝑠 𝑠ℎ𝑜𝑤𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑎𝑏𝑜𝑣𝑒. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 ∶
𝑎) 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝐻 𝑎𝑛𝑑 𝑎 6
𝑏) 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑇 𝑎𝑛𝑑 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟

Answer

𝑎) 𝑃(𝐻 𝑎𝑛𝑑 6) 𝑢𝑠𝑖𝑛𝑔 𝑎 𝑡𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚, 𝑟𝑒𝑎𝑑𝑖𝑛𝑔 𝑎𝑐𝑟𝑜𝑠𝑠 𝑤𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑤ℎ𝑖𝑙𝑠𝑡 𝑔𝑜𝑖𝑛𝑔
𝑑𝑜𝑤𝑛 𝑤𝑒 𝑠𝑢𝑚 𝑢𝑝 𝑡ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠.

1 1
𝑃(𝐻) = 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 2 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑡𝑜 𝑎 𝑐𝑜𝑖𝑛𝑡 𝑎𝑛𝑑 𝑃(6) =
2 6
1 1 1
∴ 𝑃(𝐻 𝑎𝑛𝑑 6) = 𝑃(𝐻) × 𝑃(6) = × =
2 6 12

𝑏) 𝑃(𝑇 𝑎𝑛𝑑 𝐸𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟). 𝑇ℎ𝑖𝑠 𝑡𝑖𝑚𝑒 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑙𝑜𝑜𝑘 𝑎𝑡 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒

𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑎𝑛𝑑 𝑖𝑑𝑒𝑛𝑡𝑖𝑓𝑦 𝑡ℎ𝑒 𝑜𝑛𝑒𝑠 𝑤𝑖𝑡ℎ 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟𝑠. 𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑡𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚𝑠

𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 3 𝑜𝑢𝑡 𝑜𝑓 12 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑎𝑓𝑡𝑒𝑟 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑇𝑎𝑖𝑙 𝑡ℎ𝑎𝑡 ℎ𝑎𝑣𝑒 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟𝑠

𝑤ℎ𝑖𝑐ℎ 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑏𝑒 𝑎𝑑𝑑𝑒𝑑 𝑡𝑜 𝑔𝑖𝑣𝑒 𝑡ℎ𝑒 𝑓𝑖𝑛𝑎𝑙 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦.

1 1 1
𝑃(𝑇 𝑎𝑛𝑑 𝐸𝑣𝑒𝑛) = × 𝑏𝑢𝑡 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 3 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑖𝑡 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 3 ×
2 6 12
3 1
𝑃(𝑇 𝑎𝑛𝑑 𝐸𝑣𝑒𝑛) = =
12 4

1.5.2 Example Dependent Events

A lunch box contains four sandwiches and three bananas. A person chooses at
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random an item of food and eats it. They then choose another item at random and
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Eats it. Find the probability that the first choice was a banana and the second was a
sandwich. P (B and S)

Answer:

𝐴𝑏𝑜𝑣𝑒 𝑖𝑠 𝑡ℎ𝑒 𝑡𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑠ℎ𝑜𝑤𝑖𝑛𝑔 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑣𝑒𝑛𝑡.
𝑊𝑒 𝑤𝑎𝑛𝑡 𝑡ℎ𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒 𝑜𝑓 𝑝𝑖𝑐𝑘𝑖𝑛𝑔 𝑎 𝑏𝑎𝑛𝑎𝑛𝑎 𝑡ℎ𝑒𝑛 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ.
3
𝑃(𝐵) = 𝑛𝑜𝑤 𝑛𝑜𝑡𝑒 ℎ𝑜𝑤 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑐ℎ𝑎𝑛𝑔𝑒𝑠 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑏𝑎𝑛𝑎𝑛𝑎 𝑡𝑎𝑘𝑒𝑛
7
𝑎
𝑖𝑠 𝑒𝑎𝑡𝑒𝑛. 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑠𝑝𝑎𝑐𝑒 𝑠ℎ𝑟𝑖𝑛𝑘𝑠 𝑓𝑟𝑜𝑚 7 𝑡𝑜 6 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑃(𝑋) =
7
𝑤𝑒𝑟𝑒 𝑋 = 𝑒𝑖𝑡ℎ𝑒𝑟 𝐵𝑎𝑛𝑎𝑛𝑎 𝑜𝑟 𝑆𝑎𝑛𝑑𝑤𝑖𝑐ℎ 𝑎𝑛𝑑 𝒂 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒
𝑏𝑎𝑛𝑎𝑛𝑎𝑠 𝑜𝑟 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ𝑒𝑠.

3 4 12 2
𝑃(𝐵 𝑎𝑛𝑑 𝑆) = 𝑃(𝐵) × 𝑃(𝑆) = × = =
7 6 42 7

𝑁𝑜𝑡𝑒: 𝑇ℎ𝑒 𝑜𝑟𝑑𝑒𝑟 𝑖𝑛 𝑤ℎ𝑖𝑐ℎ 𝑡ℎ𝑒 𝑒𝑣𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑛𝑒𝑒𝑑𝑒𝑑 ℎ𝑎𝑠 𝑡𝑜 𝑏𝑒 𝑡𝑎𝑘𝑒𝑛. 𝑃(𝑆 𝑎𝑛𝑑 𝐵)
𝑖𝑛 𝑡ℎ𝑖𝑠 𝑐𝑎𝑠𝑒 𝑐𝑜𝑢𝑙𝑑 ℎ𝑎𝑣𝑒 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑎𝑛𝑠𝑤𝑒𝑟 𝑏𝑢𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑤𝑟𝑜𝑛𝑔 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒, 𝑐𝑎𝑟𝑒
𝑠ℎ𝑜𝑢𝑙𝑑 𝑎𝑙𝑤𝑎𝑦𝑠 𝑏𝑒 𝑡𝑎𝑘𝑒𝑛.
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1.6.1 Exercise 1: Independent Events

1. 𝐴 𝑏𝑎𝑔 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑠 8 𝑏𝑙𝑢𝑒 𝑚𝑎𝑟𝑏𝑙𝑒𝑠 𝑎𝑛𝑑 2 𝑟𝑒𝑑 𝑚𝑎𝑟𝑏𝑙𝑒𝑠. 𝑂𝑛𝑒 𝑚𝑎𝑟𝑏𝑙𝑒 𝑖𝑠 𝑡𝑎𝑘𝑒𝑛 𝑎𝑡 𝑟𝑎𝑛𝑑𝑜𝑚
𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑑. 𝐴 𝑠𝑒𝑐𝑜𝑛𝑑 𝑚𝑎𝑟𝑏𝑙𝑒 𝑖𝑠 𝑑𝑟𝑎𝑤𝑛 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑑.
𝑎) 𝐷𝑟𝑎𝑤 𝑎 𝑡𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑡𝑜 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠.
𝑏) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑜𝑏𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑡𝑤𝑜 𝑟𝑒𝑑 𝑚𝑎𝑟𝑏𝑙𝑒𝑠?
𝑐) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑜𝑏𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑜𝑛𝑒 𝑟𝑒𝑑 𝑚𝑎𝑟𝑏𝑙𝑒 𝑎𝑛𝑑 𝑜𝑛𝑒 𝑏𝑙𝑢𝑒 𝑚𝑎𝑟𝑏𝑙𝑒?

2. 𝑆𝑒𝑎𝑛′ 𝑠 𝑙𝑢𝑛𝑐ℎ 𝑏𝑜𝑥 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑠 𝑓𝑜𝑢𝑟 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ𝑒𝑠 𝑎𝑛𝑑 𝑡ℎ𝑟𝑒𝑒 𝑏𝑎𝑛𝑎𝑛𝑎𝑠. 𝐻𝑒 𝑐ℎ𝑜𝑜𝑠𝑒𝑠 𝑎𝑛
𝑖𝑡𝑒𝑚 𝑜𝑓 𝑓𝑜𝑜𝑑 𝑎𝑛𝑑 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑠 𝑖𝑡. 𝐻𝑒 𝑡ℎ𝑒𝑛 𝑐ℎ𝑜𝑜𝑠𝑒𝑠 𝑎𝑛𝑜𝑡ℎ𝑒𝑟 𝑖𝑡𝑒𝑚 𝑎𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑎𝑛𝑑 𝑎𝑙𝑠𝑜
𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑠 𝑖𝑡.
𝑎) 𝐷𝑟𝑎𝑤 𝑡ℎ𝑒 𝑡𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠.
𝑏) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 ℎ𝑒 𝑤𝑖𝑙𝑙 𝑐ℎ𝑜𝑜𝑠𝑒 𝑎 𝑏𝑎𝑛𝑎𝑛𝑎 𝑓𝑖𝑟𝑠𝑡 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 𝑎 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ.
𝑐) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 ℎ𝑒 𝑤𝑖𝑙𝑙 𝑐ℎ𝑜𝑜𝑠𝑒 𝑎 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ 𝑓𝑖𝑟𝑠𝑡 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 𝑎 𝑏𝑎𝑛𝑎𝑛𝑎.
𝑑) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 ℎ𝑒 𝑤𝑖𝑙𝑙 𝑐ℎ𝑜𝑜𝑠𝑒 𝑡𝑤𝑜 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ𝑒𝑠.
𝑒) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 ℎ𝑒 𝑤𝑖𝑙𝑙 𝑐ℎ𝑜𝑜𝑠𝑒 𝑒𝑖𝑡ℎ𝑒𝑟 𝑎 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ 𝑜𝑟 𝑎 𝑏𝑎𝑛𝑎𝑛𝑎 𝑖𝑛 𝑎𝑛𝑦
𝑜𝑟𝑑𝑒𝑟.

3. 𝑆𝑒𝑎𝑛 𝑡ℎ𝑒𝑛 𝑑𝑒𝑐𝑖𝑑𝑒𝑠 𝑡𝑜 𝑡𝑎𝑘𝑒 𝑎𝑛𝑜𝑡ℎ𝑒𝑟 𝑖𝑡𝑒𝑚 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑙𝑢𝑛𝑐ℎ 𝑏𝑜𝑥. 𝑇ℎ𝑖𝑠 𝑚𝑒𝑎𝑛𝑠 ℎ𝑒 ℎ𝑎𝑠
𝑛𝑜𝑤 𝑚𝑎𝑑𝑒 𝑡ℎ𝑟𝑒𝑒 𝑐ℎ𝑜𝑖𝑐𝑒𝑠 𝑜𝑛𝑒 𝑎𝑓𝑡𝑒𝑟 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑤𝑖𝑡ℎ 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡.
𝑎) 𝐷𝑟𝑎𝑤 𝑡ℎ𝑒 𝑡𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑖𝑛 𝑡ℎ𝑖𝑠 𝑠𝑖𝑡𝑢𝑎𝑡𝑖𝑜𝑛.
𝑏) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 ℎ𝑒 𝑤𝑖𝑙𝑙 𝑓𝑖𝑟𝑠𝑡 𝑐ℎ𝑜𝑜𝑠𝑒 𝑎 𝑏𝑎𝑛𝑎𝑛𝑎, 𝑡ℎ𝑒𝑛 𝑎 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ 𝑎𝑛𝑑
𝑡ℎ𝑒𝑛 𝑎𝑛𝑜𝑡ℎ𝑒𝑟 𝑠𝑎𝑛𝑑𝑤𝑖𝑐ℎ.
𝑐) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 ℎ𝑒 𝑤𝑖𝑙𝑙 𝑐ℎ𝑜𝑜𝑠𝑒 𝑡ℎ𝑟𝑒𝑒 𝑏𝑎𝑛𝑎𝑛𝑎𝑠.
𝑑) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 ℎ𝑒 𝑤𝑖𝑙𝑙 𝑐ℎ𝑜𝑜𝑠𝑒 𝑡𝑤𝑜 𝑏𝑎𝑛𝑎𝑛𝑎𝑠 𝑎𝑓𝑡𝑒𝑟 ℎ𝑖𝑠 𝑡ℎ𝑖𝑟𝑑 𝑐ℎ𝑜𝑖𝑐𝑒.

4. 𝑇𝑤𝑜 𝑑𝑖𝑐𝑒 𝑎𝑟𝑒 𝑡ℎ𝑟𝑜𝑤𝑛 .

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𝑎) 𝐷𝑟𝑎𝑤 𝑎 𝑡𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑡𝑜 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠.

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𝑏) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑜𝑏𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑡𝑤𝑜 𝑛𝑢𝑚𝑏𝑒𝑟 𝑓𝑜𝑢𝑟𝑠?
𝑐) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑜𝑏𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑎 𝑡𝑤𝑜 𝑎𝑛𝑑 𝑎 𝑓𝑖𝑣𝑒?
𝑑) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑜𝑏𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑜𝑛𝑙𝑦 𝑜𝑛𝑒 𝑡𝑤𝑜?
𝑒) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑛𝑜𝑡 𝑜𝑏𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑡𝑤𝑜 𝑡ℎ𝑟𝑒𝑒𝑠 𝑙𝑎𝑛𝑑𝑖𝑛𝑔 𝑓𝑎𝑐𝑒 𝑢𝑝?

5. 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑟𝑒𝑒 𝑐𝑜𝑛𝑠𝑒𝑐𝑢𝑡𝑖𝑣𝑒 𝑠𝑜𝑐𝑐𝑒𝑟 𝑚𝑎𝑡𝑐ℎ𝑒𝑠. 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒
𝑐𝑎𝑝𝑡𝑎𝑖𝑛 𝑤𝑖𝑙𝑙 𝑤𝑖𝑛 𝑎 𝑡𝑜𝑠𝑠:
𝑖) 𝑒𝑣𝑒𝑟𝑦 𝑡𝑖𝑚𝑒?
𝑖𝑖) 𝑜𝑛𝑙𝑦 𝑜𝑛𝑐𝑒?
𝑖𝑖𝑖) 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑐𝑒?

1.6.2 Exercise 2: Dependent Events

1. 𝐴 𝑏𝑎𝑔 ℎ𝑎𝑠 6 𝑟𝑒𝑑 𝑎𝑛𝑑 4 𝑏𝑙𝑢𝑒 𝑚𝑎𝑟𝑏𝑙𝑒𝑠. 𝐴 𝑚𝑎𝑟𝑏𝑙𝑒 𝑖𝑠 𝑑𝑟𝑎𝑤𝑛 𝑏𝑢𝑡 𝑛𝑜𝑡 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑑
𝐴 𝑠𝑒𝑐𝑜𝑛𝑑 𝑚𝑎𝑟𝑏𝑙𝑒 𝑖𝑠 𝑡ℎ𝑒𝑛 𝑑𝑟𝑎𝑤𝑛 𝑎𝑛𝑑 𝑛𝑜𝑡 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑑 . 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔
𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠:
𝑎) 𝑃(𝑓𝑖𝑟𝑠𝑡 𝑚𝑎𝑟𝑏𝑙𝑒 𝑑𝑟𝑎𝑤𝑛 𝑖𝑠 𝑟𝑒𝑑)
𝑏) 𝑃(𝑏𝑜𝑡ℎ 𝑚𝑎𝑟𝑏𝑙𝑒𝑠 𝑎𝑟𝑒 𝑏𝑙𝑢𝑒)
𝑐) 𝑃(𝑜𝑛𝑒 𝑚𝑎𝑟𝑏𝑙𝑒 𝑖𝑠 𝑟𝑒𝑑 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑖𝑠 𝑏𝑙𝑢𝑒)

2. 𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑠𝑖𝑥 𝑔𝑟𝑒𝑒𝑛 𝑝𝑒𝑛𝑐𝑖𝑙𝑠 𝑎𝑛𝑑 𝑓𝑖𝑣𝑒 𝑝𝑢𝑟𝑝𝑙𝑒 𝑝𝑒𝑛𝑐𝑖𝑙𝑠 𝑜𝑛 𝑎 𝑡𝑎𝑏𝑙𝑒. 𝐽𝑎𝑚𝑒𝑠 𝑟𝑒𝑚𝑜𝑣𝑒𝑠
𝑎 𝑝𝑒𝑛𝑐𝑖𝑙 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑡𝑎𝑏𝑙𝑒 𝑎𝑛𝑑 𝑑𝑜𝑒𝑠𝑛′ 𝑡 𝑟𝑒𝑝𝑙𝑎𝑐𝑒 𝑖𝑡. 𝐻𝑒 𝑡ℎ𝑒𝑛 𝑟𝑒𝑚𝑜𝑣𝑒𝑠 𝑎𝑛𝑜𝑡ℎ𝑒𝑟 𝑝𝑒𝑛𝑐𝑖𝑙.
𝐷𝑟𝑎𝑤 𝑎 𝑡𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 . 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡:
𝑎) 𝑏𝑜𝑡ℎ 𝑝𝑒𝑛𝑐𝑖𝑙𝑠 𝑟𝑒𝑚𝑜𝑣𝑒𝑑 𝑎𝑟𝑒 𝑝𝑢𝑟𝑝𝑙𝑒
𝑏) 𝑏𝑜𝑡ℎ 𝑝𝑒𝑛𝑐𝑖𝑙𝑠 𝑟𝑒𝑚𝑜𝑣𝑒𝑑 𝑎𝑟𝑒 𝑔𝑟𝑒𝑒𝑛
𝑐) 𝑎 𝑔𝑟𝑒𝑒𝑛 𝑎𝑛𝑑 𝑝𝑢𝑟𝑝𝑙𝑒 𝑝𝑒𝑛𝑐𝑖𝑙 𝑖𝑠 𝑟𝑒𝑚𝑜𝑣𝑒𝑑.
𝑑) 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑔𝑟𝑒𝑒𝑛 𝑝𝑒𝑛𝑐𝑖𝑙 𝑖𝑠 𝑟𝑒𝑚𝑜𝑣𝑒𝑑.
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3. 𝐴 𝑏𝑎𝑔 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑠 3 𝑤ℎ𝑖𝑡𝑒 𝑚𝑎𝑟𝑏𝑙𝑒𝑠, 4 𝑏𝑙𝑢𝑒 𝑚𝑎𝑟𝑏𝑙𝑒𝑠 𝑎𝑛𝑑 3 𝑟𝑒𝑑 𝑚𝑎𝑟𝑏𝑙𝑒𝑠. 𝑇𝑤𝑜 𝑚𝑎𝑟𝑏𝑙𝑒𝑠
𝑎𝑟𝑒 𝑡𝑎𝑘𝑒𝑛 𝑜𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑔 𝑠𝑖𝑚𝑢𝑙𝑡𝑎𝑛𝑒𝑜𝑢𝑠𝑙𝑦. 𝐷𝑟𝑎𝑤 𝑎 𝑡𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑎𝑛𝑑 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒
𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡:
𝑎) 𝑏𝑜𝑡ℎ 𝑚𝑎𝑟𝑏𝑙𝑒𝑠 𝑎𝑟𝑒 𝑤ℎ𝑖𝑡𝑒 .
𝑏) 𝑏𝑜𝑡ℎ 𝑚𝑎𝑟𝑏𝑙𝑒𝑠 𝑎𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑐𝑜𝑙𝑜𝑢𝑟.
𝑐) 𝑏𝑜𝑡ℎ 𝑚𝑎𝑟𝑏𝑙𝑒𝑠 𝑑𝑖𝑓𝑓𝑒𝑟 𝑖𝑛 𝑐𝑜𝑙𝑜𝑢𝑟.

4. 𝐹𝑟𝑜𝑚 𝑎𝑛 𝑜𝑟𝑑𝑖𝑛𝑎𝑟𝑦 𝑝𝑎𝑐𝑘 𝑜𝑓 52 𝑐𝑎𝑟𝑑𝑠, 𝑡ℎ𝑒 𝑠𝑒𝑣𝑒𝑛 𝑜𝑓 𝑑𝑖𝑎𝑚𝑜𝑛𝑑𝑠 ℎ𝑎𝑠 𝑏𝑒𝑒𝑛 𝑙𝑜𝑠𝑡. 𝐴
𝑐𝑎𝑟𝑑 𝑖𝑠 𝑑𝑒𝑎𝑙𝑡 𝑓𝑟𝑜𝑚 𝑎 𝑤𝑒𝑙𝑙 𝑠ℎ𝑢𝑓𝑓𝑙𝑒𝑑 𝑑𝑒𝑐𝑘 𝑜𝑓 𝑐𝑎𝑟𝑑𝑠. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡:
𝑎) 𝑖𝑡 𝑖𝑠 𝑎 𝑑𝑖𝑎𝑚𝑜𝑛𝑑.
𝑏) 𝑎 𝑑𝑖𝑎𝑚𝑜𝑛𝑑 𝑜𝑟 𝑎 𝑞𝑢𝑒𝑒𝑛.
𝑐) 𝑎 𝑑𝑖𝑎𝑚𝑜𝑛𝑑 𝑜𝑟 𝑎 𝑠𝑒𝑣𝑒𝑛.

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1.7 Venn Diagram Problems

1.7.1 Example

𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 420 𝑙𝑒𝑎𝑟𝑛𝑒𝑟𝑠 𝑖𝑛 𝑎 𝑠𝑐ℎ𝑜𝑜𝑙. 190 𝑙𝑒𝑎𝑟𝑛𝑒𝑟𝑠 𝑝𝑙𝑎𝑦 𝑠𝑜𝑐𝑐𝑒𝑟. 260 𝑙𝑒𝑎𝑟𝑛𝑒𝑟𝑠 𝑝𝑙𝑎𝑦
𝑐𝑟𝑖𝑐𝑘𝑒𝑡 𝑎𝑛𝑑 60 𝑝𝑙𝑎𝑦 𝑏𝑜𝑡ℎ 𝑠𝑜𝑐𝑐𝑒𝑟 𝑎𝑛𝑑 𝑐𝑟𝑖𝑐𝑘𝑒𝑡.

𝑎) 𝐷𝑟𝑎𝑤 𝑎 𝑉𝑒𝑛𝑛 𝐷𝑖𝑎𝑔𝑟𝑎𝑚 𝑡𝑜 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑡ℎ𝑖𝑠 𝑖𝑛𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛

𝑏) 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑎 𝑙𝑒𝑎𝑟𝑛𝑒𝑟 𝑐ℎ𝑜𝑠𝑒𝑛 𝑎𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑝𝑙𝑎𝑦𝑠 𝑛𝑒𝑖𝑡ℎ𝑒𝑟

𝑠𝑜𝑐𝑐𝑒𝑟 𝑛𝑜𝑟 𝑐𝑟𝑖𝑐𝑘𝑒𝑡.
𝑐) 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑎 𝑙𝑒𝑎𝑟𝑛𝑒𝑟 𝑐ℎ𝑜𝑠𝑒𝑛 𝑎𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑝𝑙𝑎𝑦𝑠 𝑛𝑒𝑖𝑡ℎ𝑒𝑟
𝑠𝑜𝑐𝑐𝑒𝑟 𝑛𝑜𝑟 𝑐𝑟𝑖𝑐𝑘𝑒𝑡.

𝑑) 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑎 𝑙𝑒𝑎𝑟𝑛𝑒𝑟 𝑐ℎ𝑜𝑠𝑒𝑛 𝑎𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑝𝑙𝑎𝑦𝑠 𝑜𝑛𝑙𝑦 𝑜𝑛𝑒 𝑜𝑓
𝑡ℎ𝑒 𝑡𝑤𝑜 𝑠𝑝𝑜𝑟𝑡𝑠.

𝑒) 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑎 𝑙𝑒𝑎𝑟𝑛𝑒𝑟 𝑐ℎ𝑜𝑠𝑒𝑛 𝑎𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑝𝑙𝑎𝑦
𝑏𝑜𝑡ℎ 𝑠𝑝𝑜𝑟𝑡𝑠.

𝑓) 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑎 𝑙𝑒𝑎𝑟𝑛𝑒𝑟 𝑐ℎ𝑜𝑠𝑒𝑛 𝑎𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑝𝑙𝑎𝑦𝑠

𝑏𝑜𝑡ℎ 𝑠𝑝𝑜𝑟𝑡𝑠.

Answer

11
Page
 130 + 60 + 200 390
𝑏) 𝑃(𝑆 𝑜𝑟 𝐶) = = = 0.92
420 420
30
𝑐) 𝑃(𝑛𝑜𝑡 𝑆 𝑜𝑟 𝑛𝑜𝑡 𝐶) = = 0.07
420
130
𝑑) 𝑃(𝑜𝑛𝑙𝑦 𝑜𝑛𝑒 𝑠𝑝𝑜𝑟𝑡) = = 0.31
420
130 200 30 360
𝑒) 𝑃(𝑛𝑜𝑡 𝑝𝑙𝑎𝑦𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑝𝑜𝑟𝑡𝑠) = + + = = 0.86
420 420 420 420
60 1
𝑓) 𝑃(𝑆 𝑎𝑛𝑑 𝐶) = =
420 7

1.7.2 Exercise 3: Venn diagram Problems

1. 𝐼𝑡 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝑃(𝐴) = 0.35 , 𝑃(𝐵) = 0.8 𝑎𝑛𝑑 𝑃(𝐴 𝑎𝑛𝑑 𝐵) = 0.25.
𝑈𝑠𝑒 𝑎 𝑉𝑒𝑛𝑛 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑎𝑛𝑑 𝑎 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑡𝑜 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒:
𝑎) 𝑃(𝐴 𝑜𝑟 𝐵)
𝑏) 𝑃(𝐴′ 𝑎𝑛𝑑 𝐵)
𝑐) 𝑃(𝐴′ 𝑜𝑟 𝐵)
𝑑) 𝑃(𝐴 𝑎𝑛𝑑 𝐵)
𝑒) 𝑃(𝐴 𝑜𝑟 𝐵)′

Note the notation 𝑨′ 𝒊𝒎𝒑𝒍𝒊𝒆𝒔 𝒄𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕 𝒐𝒇 𝑨

2. 𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 130 𝐺𝑟𝑎𝑑𝑒 11 𝑙𝑒𝑎𝑟𝑛𝑒𝑟𝑠 𝑎𝑡 𝑎 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑠𝑐ℎ𝑜𝑜𝑙. 50 𝑙𝑒𝑎𝑟𝑛𝑒𝑟𝑠 𝑡𝑎𝑘𝑒

𝑏𝑖𝑜𝑙𝑜𝑔𝑦 𝑎𝑛𝑑 68 𝑡𝑎𝑘𝑒 𝑀𝑎𝑡ℎ𝑠. 32 𝑙𝑒𝑎𝑟𝑛𝑒𝑟𝑠 𝑡𝑎𝑘𝑒 𝑏𝑜𝑡ℎ 𝑠𝑢𝑏𝑗𝑒𝑐𝑡𝑠.

𝑎) 𝐷𝑟𝑎𝑤 𝑎 𝑉𝑒𝑛𝑛 𝐷𝑖𝑎𝑔𝑟𝑎𝑚 𝑡𝑜 𝑖𝑙𝑙𝑢𝑠𝑡𝑟𝑎𝑡𝑒 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑑𝑎𝑡𝑎.

𝑏) 𝐻𝑒𝑛𝑐𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑎 𝑙𝑒𝑎𝑟𝑛𝑒𝑟 𝑖𝑛 𝐺𝑟𝑎𝑑𝑒 11 𝑐ℎ𝑜𝑠𝑒𝑛 𝑎𝑡
12

𝑟𝑎𝑛𝑑𝑜𝑚 ∶ 𝑖) 𝑇𝑎𝑘𝑒𝑠 𝐵𝑖𝑜𝑙𝑜𝑔𝑦 𝑖𝑖) 𝑇𝑎𝑘𝑒𝑠 𝑀𝑎𝑡ℎ𝑠 𝑎𝑛𝑑 𝐵𝑖𝑜𝑙𝑜𝑔𝑦

Page
3. 𝐴 𝑔𝑟𝑜𝑢𝑝 𝑜𝑓 80 𝑎𝑡ℎ𝑙𝑒𝑡𝑒𝑠 𝑒𝑛𝑡𝑒𝑟𝑒𝑑 𝑎 100𝑚, 200𝑚 𝑎𝑛𝑑 400𝑚 𝑠𝑝𝑟𝑖𝑛𝑡𝑠 𝑎𝑠 𝑓𝑜𝑙𝑙𝑜𝑤𝑠
6 𝑒𝑛𝑡𝑒𝑟𝑒𝑑 𝑎𝑙𝑙 𝑡ℎ𝑟𝑒𝑒 𝑒𝑣𝑒𝑛𝑡𝑠
21 𝑒𝑛𝑡𝑒𝑟𝑒𝑑 𝑛𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑣𝑒𝑛𝑡𝑠
10 𝑒𝑛𝑡𝑒𝑟𝑒𝑑 𝑡ℎ𝑒 100𝑚 𝑎𝑛𝑑 200𝑚 𝑒𝑣𝑒𝑛𝑡𝑠
11 𝑒𝑛𝑡𝑒𝑟𝑒𝑑 𝑡ℎ𝑒 200𝑚 𝑎𝑛𝑑 400𝑚 𝑒𝑣𝑒𝑛𝑡𝑠
𝑜𝑓 𝑡ℎ𝑒 21 𝑤ℎ𝑜 𝑒𝑛𝑡𝑒𝑟𝑒𝑑 𝑡ℎ𝑒 100𝑚 𝑒𝑣𝑒𝑛𝑡, 10 𝑒𝑛𝑡𝑒𝑟𝑒𝑑 𝑛𝑜𝑡ℎ𝑖𝑛𝑔 𝑒𝑙𝑠𝑒.
27 𝑒𝑛𝑡𝑒𝑟𝑒𝑑 𝑡ℎ𝑒 400𝑚.

𝑎) 𝑅𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑠𝑖𝑡𝑢𝑎𝑡𝑖𝑜𝑛 𝑢𝑠𝑖𝑛𝑔 𝑎 𝑉𝑒𝑛𝑛 𝐷𝑖𝑎𝑔𝑟𝑎𝑚.

𝑏) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑎𝑡ℎ𝑙𝑒𝑡𝑒𝑠 𝑒𝑛𝑡𝑒𝑟𝑒 𝑡ℎ𝑒 200𝑚 𝑒𝑣𝑒𝑛?
𝑐)𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑎𝑛 𝑎𝑡ℎ𝑙𝑒𝑡𝑒, 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑 𝑎𝑡 𝑟𝑎𝑛𝑑𝑜𝑚, 𝑟𝑢𝑛𝑛𝑖𝑛𝑔 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑡𝑤𝑜
𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝𝑟𝑖𝑛𝑡 𝑒𝑣𝑒𝑛𝑡𝑠?

4. 𝐴 𝑔𝑟𝑜𝑢𝑝 𝑜𝑓 50 𝑝𝑒𝑜𝑝𝑙𝑒 𝑤𝑎𝑠 𝑎𝑠𝑘𝑒𝑑 𝑤ℎ𝑖𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑡ℎ𝑟𝑒𝑒 𝑛𝑒𝑤𝑠𝑝𝑎𝑝𝑒𝑟𝑠 𝐴, 𝐵 𝑜𝑟 𝐶 𝑡ℎ𝑒𝑦

𝑟𝑒𝑎𝑑. 𝑇ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑠 𝑠ℎ𝑜𝑤𝑒𝑑 𝑡ℎ𝑎𝑡 25 𝑟𝑒𝑎𝑑 𝐴, 16 𝑟𝑒𝑎𝑑 𝐵, 14 𝑟𝑒𝑎𝑑 𝐶, 5 𝑟𝑒𝑎𝑑 𝑏𝑜𝑡ℎ 𝐴 𝑎𝑛𝑑 𝐵
4 𝑟𝑒𝑎𝑑 𝑏𝑜𝑡ℎ 𝐵 𝑎𝑛𝑑 𝐶, 6 𝑟𝑒𝑎𝑑 𝑏𝑜𝑡ℎ 𝐶 𝑎𝑛𝑑 𝐴 𝑎𝑛𝑑 2 𝑟𝑒𝑎𝑑 𝑎𝑙𝑙 3.
𝑎) 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑡ℎ𝑖𝑠 𝑖𝑛𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑛 𝑎 𝑉𝑒𝑛𝑛 𝐷𝑖𝑎𝑔𝑟𝑎𝑚.
𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑎 𝑝𝑒𝑟𝑠𝑜𝑛 𝑐ℎ𝑜𝑠𝑒𝑛 𝑎𝑡 𝑟𝑎𝑛𝑑𝑜𝑚 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑔𝑟𝑜𝑢𝑝 𝑟𝑒𝑎𝑑𝑠
𝑏) 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 1 𝑜𝑓 𝑡ℎ𝑒 𝑛𝑒𝑤𝑠𝑝𝑎𝑝𝑒𝑟𝑠
𝑐)𝑜𝑛𝑙𝑦 1 𝑜𝑓 𝑡ℎ𝑒 𝑛𝑒𝑤𝑠𝑝𝑎𝑝𝑒𝑟𝑠
𝑑) 𝑜𝑛𝑙𝑦 𝐴.
13Page
1.8 Contingent Tables

1.8.1 Example

A Durban gym recorded the number of members (men and women) who use
or do not use the gym on a regular basis. The results are recorded in the
following contingency table.

Men Women Total

Uses the gym 70 150 220
regularly
Don’t use the gym 110 40 150
regularly
Total 180 190 370

Calculate the probability that a member selected at random from the sample
of 370 members:

a) Uses the gym regularly.

b) Uses the gym regularly given the member is a woman.
c) Does not use the gym regularly given the member is a man.

Answer:

220
𝑎) 𝑃(𝑔𝑦𝑚 𝑟𝑒𝑔𝑢𝑙𝑎𝑟𝑙𝑦) = = 0.59
370

150
𝑏) 𝑃(𝑔𝑦𝑚 𝑟𝑒𝑔𝑢𝑙𝑎𝑟 𝑔𝑖𝑣𝑒𝑛 𝑖𝑡𝑠 𝑎 𝑤𝑜𝑚𝑎𝑛) = = 0.79
190
14

110
Page

𝑐) 𝑃(𝑁𝑜𝑡 𝑟𝑒𝑔𝑢𝑙𝑎𝑟 𝑎𝑛𝑑 𝑖𝑠 𝑎 𝑚𝑎𝑛) = = 0.61

180
1.8.2 Exercise 4: Contingent Tables

15
Page
1.9 The Counting Principle

Probability problems dealt with so far address the likelihood of an event

happening. Often the possibility space is given as in picking from a deck of 52
cards or tossing a coin (which has Heads or Tails). In some instances, it is not
clearly defined and hence we need to first establish the probability space. An
example would be students in a class and they have to seat in a row of 10 chairs. If
the exercise is done such that 1 student is picked from the group each time at
random, we see that the possibility space gradually changes each time, that is there
are 10 possible students for the first spot, 9 possible students for the second, 8 for
the third and so forth. This principle of identifying the possible ways in which an
outcome can be obtained is the Counting Principle

If task 1 can be performed in n ways and task 2 can be performed in m ways,

then there are n x m number of ways in which both tasks can be completed.

There are two types of problems where the counting principle can be applied. The
first is when there is no repetition, the second is in which repetition is allowed.

Example 1: Repetition not allowed

𝑃𝑒𝑟𝑠𝑜𝑛 𝐴 ℎ𝑎𝑠 𝑡𝑜 𝑝𝑖𝑐𝑘 𝑎𝑛 𝑎𝑡𝑡𝑖𝑟𝑒 𝑡𝑜 𝑤𝑒𝑎𝑟 𝑓𝑜𝑟 𝑎 𝑤𝑒𝑑𝑑𝑖𝑛𝑔. 𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 3 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑠ℎ𝑜𝑒𝑠
5 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑠ℎ𝑖𝑟𝑡𝑠 𝑎𝑛𝑑 4 𝑝𝑎𝑖𝑟𝑠 𝑜𝑓 𝑡𝑖𝑒𝑠. 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠 𝑐𝑎𝑛 𝑏𝑒
𝑐ℎ𝑜𝑠𝑒𝑛?

𝐴𝑛𝑠𝑤𝑒𝑟:
𝐼𝑛 𝑡ℎ𝑖𝑠 𝑠𝑐𝑒𝑛𝑎𝑟𝑖𝑜, 𝐴 𝑐𝑎𝑛 𝑜𝑛𝑙𝑦 𝑝𝑖𝑐𝑘 1 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑡𝑒𝑚𝑠 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑙𝑖𝑠𝑡 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒𝑦 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒
𝑟𝑒𝑝𝑒𝑎𝑡𝑒𝑑. 𝑆𝑜 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑐𝑎𝑡𝑒𝑔𝑜𝑟𝑦, 𝐴 𝑤𝑖𝑙𝑙 𝑝𝑖𝑐𝑘 1 𝑠ℎ𝑜𝑒, 1 𝑠ℎ𝑖𝑟𝑡 𝑎𝑛𝑑 1 𝑡𝑖𝑒, 𝑜𝑢𝑡 𝑜𝑓 𝑎
16

𝑝𝑜𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 3 𝑠ℎ𝑜𝑒𝑠, 5 𝑠ℎ𝑖𝑟𝑡𝑠 𝑎𝑛𝑑 4 𝑡𝑖𝑒𝑠 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦, ℎ𝑒𝑛𝑐𝑒:

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𝑇𝑎𝑠𝑘 1 = 3 𝑤𝑎𝑦𝑠
𝑇𝑎𝑠𝑘 2 = 5 𝑤𝑎𝑦𝑠
𝑇𝑎𝑠𝑘 3 = 4 𝑤𝑎𝑦𝑠

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 3 × 5 × 4 = 60 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠

Example 2: When Repetition is allowed

𝐼𝑛 𝑎 𝑔𝑖𝑣𝑒𝑛 𝑠𝑐𝑒𝑛𝑎𝑟𝑖𝑜, 𝑝𝑒𝑟𝑠𝑜𝑛 𝐴 𝑖𝑠 𝑛𝑜𝑤 𝑎𝑠𝑘𝑒𝑑 𝑡𝑜 𝑝𝑖𝑐𝑘 𝑎 4 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑓𝑟𝑜𝑚

𝑎 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 10 𝑑𝑖𝑔𝑖𝑡𝑠 0 − 9 𝑖𝑛𝑐𝑙𝑢𝑠𝑖𝑣𝑒. 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠
𝑎𝑟𝑒 𝑡ℎ𝑒𝑟𝑒?
𝑨𝒏𝒔𝒘𝒆𝒓:
𝑆𝑖𝑛𝑐𝑒 𝑒𝑎𝑐ℎ 𝑡𝑎𝑠𝑘 𝑡𝑜 𝑏𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒𝑑 𝑖𝑠 𝑡𝑎𝑘𝑖𝑛𝑔 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑓𝑟𝑜𝑚 𝑎 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 10
𝑖𝑡 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 10 × 10 × 10 × 10 = 10 000 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠.

Factorial Notation

Given a word MOUSE, how many possible ways can these letters be arranged
without repetition? Since it’s a five letter word, it follows there are 5 x 4 x 3 x 2 x
1 ways of arranging the positions of each letter. This is known as a pure
arrangement

𝑻𝒉𝒆 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒇𝒂𝒄𝒕𝒐𝒓𝒊𝒂𝒍 𝒏𝒐𝒕𝒂𝒕𝒊𝒐𝒏 𝒊𝒔 𝒔𝒖𝒄𝒉 𝒕𝒉𝒂𝒕 𝒈𝒊𝒗𝒆𝒏 𝒏 𝒕𝒆𝒓𝒎𝒔:
𝒏! = 𝒏(𝒏 − 𝟏)(𝒏 − 𝟐) … (𝟏)

The process involves multiplying numbers that are reduced by 1 consecutively

until we reach 1.
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Restricted Positions

In some instances there maybe restrictions on the positions. In the previous

example, if the restriction was such that the first letter has to be M, then it changes
the entire problem, as now it is no longer a pure arrangement.

Probability

We can use the counting Principle to find probability of a required outcome/event.

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑚𝑒𝑒𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛

𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 (𝐸𝑣𝑒𝑛𝑡) =
𝑇𝑜𝑡𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠

Example 3: Probability

Five men and four women are seated on a bench together. What is the probability
that the four women will be seated together?

𝑨𝒏𝒔𝒘𝒆𝒓:

𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑜𝑓 𝑝𝑒𝑜𝑝𝑙𝑒 𝑠𝑒𝑎𝑡𝑖𝑛𝑔 𝑜𝑛 𝑎 𝑏𝑒𝑛𝑐ℎ 𝑖𝑛 9!

𝐼𝑓 𝑡ℎ𝑒 𝑤𝑜𝑚𝑒𝑛 𝑤𝑒𝑟𝑒 𝑠𝑒𝑎𝑡𝑒𝑑 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟 𝑡ℎ𝑒 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 4! × 6!
𝑡ℎ𝑖𝑠 𝑖𝑠 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 4 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠 𝑜𝑓 𝑠𝑖𝑡𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑤𝑜𝑚𝑒𝑛, 𝑎𝑛𝑑 𝑜𝑛𝑐𝑒
𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑜𝑛𝑒 𝑖𝑠 𝑠𝑒𝑎𝑡𝑒𝑑, 𝑦𝑜𝑢 𝑎𝑟𝑒 𝑙𝑒𝑓𝑡 𝑤𝑖𝑡ℎ 6 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑠 𝑡𝑜 𝑓𝑖𝑙𝑙.
4! × 6! 1
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑏𝑒𝑐𝑜𝑚𝑒𝑠: =
9! 21

1.9.1 Exercise 5: Counting Principle

1. 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑒 𝑤𝑜𝑟𝑑 𝑃𝑅𝑂𝐵𝐴𝐵𝐼𝐿𝐼𝑇𝑌 (𝑅𝑒𝑝𝑒𝑎𝑡𝑖𝑛𝑔 𝑙𝑒𝑡𝑡𝑒𝑟𝑠 𝑎𝑟𝑒 𝑟𝑒𝑔𝑎𝑟𝑑𝑒𝑑 𝑎𝑠 𝑖𝑑𝑒𝑛𝑡𝑖𝑐𝑎𝑙)

𝑎) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑒𝑡𝑡𝑒𝑟𝑠 𝑜𝑓 𝑡ℎ𝑖𝑠 𝑤𝑜𝑟𝑑 𝑎𝑟𝑒

18

𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒.
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𝑏) 𝑖𝑓 𝑡ℎ𝑒 𝑙𝑒𝑡𝑡𝑒𝑟𝑠 𝑜𝑓 𝑡ℎ𝑖𝑠 𝑤𝑜𝑟𝑑 𝑎𝑟𝑒 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑑 𝑟𝑎𝑛𝑑𝑜𝑚𝑙𝑦, 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦
𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔 𝑤𝑜𝑟𝑑 𝑤𝑖𝑙𝑙 𝑠𝑡𝑎𝑟𝑡 𝑎𝑛𝑑 𝑒𝑛𝑑 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 𝑙𝑒𝑡𝑡𝑒𝑟 𝐵.

2. 𝐹𝑖𝑣𝑒 𝑙𝑒𝑎𝑟𝑛𝑒𝑟𝑠 (𝑡ℎ𝑟𝑒𝑒 𝑏𝑜𝑦𝑠 𝑎𝑛𝑑 𝑡𝑤𝑜 𝑔𝑖𝑟𝑙𝑠) 𝑎𝑟𝑒 𝑡𝑜 𝑏𝑒 𝑠𝑒𝑎𝑡𝑒𝑑 𝑖𝑛 𝑎 𝑟𝑜𝑤 𝑓𝑜𝑟
𝑎 𝑝ℎ𝑜𝑡𝑜𝑔𝑟𝑎𝑝ℎ.

𝑎) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑤𝑎𝑦𝑠 𝑐𝑎𝑛 𝑡ℎ𝑒𝑠𝑒 𝑙𝑒𝑎𝑟𝑛𝑒𝑟𝑠 𝑎𝑟𝑒 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑑?

𝑏) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑤𝑎𝑦𝑠 𝑐𝑎𝑛 𝑡ℎ𝑒𝑦 𝑏𝑒 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑑 𝑖𝑓 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑔𝑖𝑟𝑙𝑠 𝑟𝑒𝑓𝑢𝑠𝑒 𝑡𝑜 𝑠𝑖𝑡 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟?
𝑐) 𝐼𝑓 𝑡ℎ𝑒 𝑙𝑒𝑎𝑟𝑛𝑒𝑟𝑠 𝑎𝑟𝑒 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑑 𝑟𝑎𝑛𝑑𝑜𝑚𝑙𝑦, 𝑤ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜
𝑔𝑖𝑟𝑙𝑠 𝑤𝑖𝑙𝑙 𝑛𝑜𝑡 𝑠𝑖𝑡 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟?

3. 𝑇ℎ𝑒 𝑑𝑖𝑔𝑖𝑡𝑠 3,4,6,7,8 𝑎𝑛𝑑 9 𝑎𝑟𝑒 𝑢𝑠𝑒𝑑 𝑡𝑜 𝑐𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡 𝑎 𝑓𝑜𝑢𝑟 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟.
𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑎𝑟𝑒 𝑡ℎ𝑒𝑟𝑒 𝑖𝑓:
𝑎) 𝑟𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑𝑖𝑔𝑖𝑡𝑠 𝑖𝑠 𝑎𝑙𝑙𝑜𝑤𝑒𝑑.
𝑏) 𝑟𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑𝑖𝑔𝑖𝑡𝑠 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑙𝑙𝑜𝑤𝑒𝑑.
𝑐) 𝑟𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑𝑖𝑔𝑖𝑡𝑠 𝑖𝑠 𝑎𝑙𝑙𝑜𝑤𝑒𝑑 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 2.
𝑑) 𝑟𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑑𝑖𝑔𝑖𝑡𝑠 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑙𝑙𝑜𝑤𝑒𝑑 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 2.

4. 𝑇ℎ𝑒 𝑙𝑒𝑡𝑡𝑒𝑟𝑠 𝑖𝑓 𝑡ℎ𝑒 𝑤𝑜𝑟𝑑 𝑀𝐴𝑇𝑅𝐼𝑋 𝑎𝑟𝑒 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑑 𝑟𝑎𝑛𝑑𝑜𝑚𝑙𝑦.

𝑎) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑡ℎ𝑒𝑟𝑒?
𝑏) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑙𝑎𝑠𝑡 𝑙𝑒𝑡𝑡𝑒𝑟 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑎 𝑣𝑜𝑤𝑒𝑙?
𝑐) 𝑊ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑣𝑜𝑤𝑒𝑙𝑠 𝑤𝑖𝑙𝑙 𝑎𝑝𝑝𝑒𝑎𝑟 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟?

5. 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑒 𝑤𝑜𝑟𝑑 𝐶𝐻𝐸𝑀𝐼𝑆𝑇𝑅𝑌.

𝑎) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑤𝑎𝑦𝑠 𝑐𝑎𝑛 𝑡ℎ𝑒 𝑙𝑒𝑡𝑡𝑒𝑟𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑜𝑟𝑑 𝑏𝑒 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑑?
𝑏) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑚𝑎𝑑𝑒 𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 𝑎 𝐶 𝑎𝑛𝑑 𝑒𝑛𝑑𝑖𝑛𝑔
𝑤𝑖𝑡ℎ 𝑎 𝑌?
𝑐) 𝐻𝑜𝑤 𝑚𝑎𝑛𝑦 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑚𝑎𝑑𝑒 𝑖𝑓 𝑡ℎ𝑒 𝑙𝑒𝑡𝑡𝑒𝑟𝑠 𝑀, 𝐼 𝑎𝑛𝑑 𝑆 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑔𝑟𝑜𝑢𝑝𝑒𝑑
19

𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟 𝑖𝑛 𝑎𝑛𝑦 𝑜𝑟𝑑𝑒𝑟?

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1.10 Exam Problems: Probability

20
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21
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22
 Solutions

1.11.1 Exercise 1

1𝑎) 𝑇𝑟𝑒𝑒 𝐷𝑖𝑎𝑔𝑟𝑎𝑚 ∶

2 2 4
𝑏) 𝑃(𝑅𝑅) = × =
10 10 10

𝑐) 𝑃(𝑜𝑛𝑒 𝑅𝑒𝑑 𝑎𝑛𝑑 𝑜𝑛𝑒 𝐵𝑙𝑢𝑒) = 𝑃(𝑅𝐵) + 𝑃(𝐵𝑅)

𝑡ℎ𝑖𝑠 𝑖𝑠 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑡𝑤𝑜 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑜𝑓 𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑜𝑛𝑒 𝑟𝑒𝑑 𝑎𝑛𝑑 𝑜𝑛𝑒 𝑏𝑙𝑢𝑒
8 2 32
∴ 𝑖𝑡 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 × ×2= = 0.32.
10 10 100

2. 𝑇𝑟𝑒𝑒 𝐷𝑖𝑎𝑔𝑟𝑎𝑚

23
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 3 4 12
𝑏) 𝑃(𝐵𝑎𝑛𝑎𝑛𝑎 𝑡ℎ𝑒𝑛 𝑆𝑎𝑛𝑑𝑤𝑖𝑐ℎ) = 𝑃(𝐵𝑆)𝑜𝑛 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 = × =
7 7 49
3 4 12
𝑐) 𝑃(𝑆𝑎𝑛𝑑𝑤𝑖𝑐ℎ 𝑡ℎ𝑒𝑛 𝐵𝑎𝑛𝑎𝑛𝑎) = 𝑃(𝑆𝐵) = × =
7 7 49
4 4 16
𝑑) 𝑃(𝑡𝑤𝑜 𝑆𝑎𝑛𝑑𝑤𝑖𝑐ℎ𝑒𝑠) = 𝑃(𝑆𝑆) = × =
7 7 49
12 12
𝑒) 𝑃(𝑆𝑎𝑛𝑑𝑤𝑖𝑐ℎ 𝑜𝑟 𝐵𝑎𝑛𝑎𝑛𝑎) = 𝑃(𝑆𝐵) + 𝑃(𝐵𝑆)𝑡ℎ𝑒 𝑡𝑤𝑜 𝑝𝑜𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 = +
49 49
24
𝑃(𝐵𝑆) + 𝑃(𝑆𝐵) =
49

3. 𝑇𝑟𝑒𝑒 𝐷𝑖𝑎𝑔𝑟𝑎𝑚

3 4 4 48
𝑏) 𝑃(𝑏𝑎𝑛𝑎𝑛𝑎 𝑡ℎ𝑒𝑛 𝑆𝑎𝑛𝑑𝑤𝑖𝑐ℎ 𝑡ℎ𝑒𝑛 𝑆𝑎𝑛𝑑𝑤𝑖𝑐ℎ) = 𝑃(𝐵𝑆𝑆) = × × =
7 7 7 343
3 3 3 27
𝑐) 𝑃(𝑡ℎ𝑟𝑒𝑒 𝑏𝑎𝑛𝑎𝑛𝑎𝑠) = 𝑃(𝐵𝐵𝐵) = × × =
7 7 7 343
24

3 3 4
𝑑) 𝑃(2 𝑏𝑎𝑛𝑎𝑛𝑎𝑠 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑑) = 𝑃(𝐵𝐵𝑆) + 𝑃(𝐵𝑆𝐵) + 𝑃(𝑆𝐵𝐵) = × × ×3
7 7 7
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 36 108
𝑃(2 𝑏𝑎𝑛𝑎𝑛𝑎𝑠) = ×3=
343 343

4. 𝑇𝑟𝑒𝑒 𝐷𝑖𝑎𝑔𝑟𝑎𝑚 𝑓𝑜𝑟 𝑡𝑤𝑜 𝑑𝑖𝑐𝑒

1 1 1
𝑏) 𝑃(4 𝑎𝑛𝑑 4) = × =
6 6 36
1 1 2
𝑐) 𝑃(𝑔𝑒𝑡𝑡𝑖𝑛𝑔 2 𝑎𝑛𝑑 𝑎 5) = 𝑃(2 𝑎𝑛𝑑 5) + 𝑃(5 𝑎𝑛𝑑 2) = × ×2=
6 6 36

𝑑) 𝑃(𝑜𝑛𝑙𝑦 𝑜𝑛𝑒 2) = 𝑃(2 𝑎𝑛𝑑 𝐷𝑖𝑓𝑓 𝑁𝑢𝑚𝑏𝑒𝑟) + 𝑃(𝐷𝑖𝑓𝑓 𝑁𝑢𝑚𝑏𝑒𝑟 𝑎𝑛𝑑 2)

1 5 5 1 10
= × + × =
6 6 6 6 36

1 1 35
𝑒) 𝑃(𝑛𝑜 𝑡𝑤𝑜 3𝑠 𝑓𝑎𝑐𝑖𝑛𝑔 𝑢𝑝) = 𝑃(33′ ) = 1 − 𝑃(33) = 1 − × =
6 6 36
1 1 1 1
5. 𝑃(𝑇 𝑒𝑣𝑒𝑟𝑦𝑡𝑖𝑚𝑒) = 𝑃(𝑇𝑇𝑇) = × × =
2 2 2 8
25

1 3
𝑏) 𝑃(𝑇 𝑜𝑛𝑐𝑒) = 𝑃(𝑇𝑇 ′ 𝑇 ′ ) + 𝑃(𝑇 ′ 𝑇 ′ 𝑇) + 𝑃(𝑇 ′ 𝑇𝑇 ′ ) = ×3=
Page

8 8
 1
𝑐) 𝑃(𝑇 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑐𝑒): 𝐹𝑖𝑟𝑠𝑡 𝑃(𝑇 𝑛𝑒𝑣𝑒𝑟) = 𝑃(𝑇 ′ 𝑇 ′ 𝑇 ′ ) =
8
1 7
∴ 𝑃(𝑇 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑐𝑒) = 1 − =
8 8

1.11.2 Exercise 2

6
1𝑎) 𝑃(𝑓𝑖𝑟𝑠𝑡 𝑚𝑎𝑟𝑏𝑙𝑒 𝑑𝑟𝑎𝑤𝑛 𝑖𝑠 𝑟𝑒𝑑) = 𝑃(𝑅) =
10
4 3 12 2
𝑏) 𝑃(𝑏𝑜𝑡ℎ 𝑚𝑎𝑟𝑏𝑙𝑒𝑠 𝑎𝑟𝑒 𝑏𝑙𝑢𝑒) = 𝑃(𝐵𝐵) = × = =
10 9 90 15
𝑐) 𝑃(𝑜𝑛𝑒 𝑚𝑎𝑟𝑏𝑙𝑒 𝑖𝑠 𝑟𝑒𝑑 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑖𝑠 𝑏𝑙𝑢𝑒) = 𝑃(𝑅𝐵) + 𝑃(𝐵𝑅)
6 4 48 8
= × ×2= =
10 9 90 15

2. 6 𝐺𝑅𝐸𝐸𝑁 𝑎𝑛𝑑 5 𝑃𝑢𝑟𝑝𝑙𝑒 𝑡𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚

5 4 20 2
𝑎) 𝑃(𝑏𝑜𝑡ℎ 𝑃𝑢𝑟𝑝𝑙𝑒) = 𝑃(𝑃𝑃) = × = =
11 10 110 11

6 5 30 3
𝑏) 𝑃(𝐺𝐺) = × = =
11 10 110 11
26

6 5 5 6 60 6
𝑐) 𝑃(𝐺𝑃)𝑜𝑟 𝑃(𝑃𝐺) = × + × = =
Page

11 10 11 10 110 11
 5 4 2 2 9
𝑑) 𝑃(𝐺 ′ 𝐺 ′ ) = × = 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑔𝑟𝑒𝑒𝑛) = 1 − =
11 10 11 11 11

3. 𝑇𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑙𝑒𝑓𝑡 𝑎𝑠 𝑒𝑥𝑒𝑟𝑐𝑖𝑠𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑡𝑢𝑑𝑒𝑛𝑡: 𝑁𝑜𝑡𝑒 𝑖𝑡 ℎ𝑎𝑠 3 𝑏𝑟𝑎𝑛𝑐ℎ𝑒𝑠

3 2 6 1
𝑎) 𝑃(𝑊𝑊) = × = =
10 9 90 15
1 4 3 3 2 24 4
𝑏) 𝑃(𝑊𝑊)𝑜𝑟 𝑃(𝐵𝐵)𝑜𝑟 𝑃(𝑅𝑅) = + × + × = =
15 10 9 10 9 90 15
4 11
𝑐) 𝑃(𝑏𝑜𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟 𝑖𝑛 𝑐𝑜𝑙𝑜𝑢𝑟) = 1 − 𝑃(𝑠𝑎𝑚𝑒 𝑐𝑜𝑙𝑜𝑢𝑟) = 1 − =
15 15
12
4 𝑎) 𝑃(𝐷𝑖𝑎𝑚𝑜𝑛𝑑) =
51

12 4 1 15
𝑏) 𝑃(𝐷𝑖𝑎𝑚𝑜𝑛𝑑 𝑜𝑟 𝑄𝑢𝑒𝑒𝑛) = + − =
51 51 51 51
12 4 16
𝑐) 𝑃(𝐷𝑖𝑎𝑚𝑜𝑛𝑑 𝑜𝑟 7) = + =
51 51 51

1.11.3 Exercise 3

27

𝑎) 𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴𝑎𝑛𝑑𝐵) = 0.35 + 0.8 − 0.25 = 0.9

Page

𝑏) 𝑃(𝐴′ 𝑎𝑛𝑑 𝐵) = 0.55 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑣𝑒𝑛𝑛 𝑑𝑖𝑎𝑔𝑟𝑎𝑚

𝑐) 𝑃(𝐴′ 𝑜𝑟 𝐵) = 𝑃(𝐴′ ) + 𝑃(𝐵) − 𝑃(𝐴′ 𝑎𝑛𝑑 𝐵) = 0.65 + 0.8 − 0.55 = 0.9
𝑑) 𝑃(𝐴 𝑎𝑛𝑑 𝐵)′ = 0.1 + 0.55 + 0.1 = 0.75
𝑒) 𝑃(𝐴 𝑜𝑟 𝐵)′ = 0.1 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑣𝑒𝑛𝑛 𝑑𝑖𝑎𝑔𝑟𝑎𝑚

2. 𝑉𝑒𝑛𝑛 𝐷𝑖𝑎𝑔𝑟𝑎𝑚 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑐𝑙𝑎𝑠𝑠

50 5
2𝑖) 𝑃(𝐵𝑖𝑜𝑙𝑜𝑔𝑦) = =
130 13
32 16
2𝑖𝑖) 𝑃(𝑀𝑎𝑡ℎ𝑠 𝑎𝑛𝑑 𝐵𝑖𝑜𝑙𝑜𝑔𝑦) = =
130 65

3. 𝑉𝑒𝑛𝑛 𝐷𝑖𝑎𝑔𝑟𝑎𝑚 𝑓𝑜𝑟 𝐸𝑣𝑒𝑛𝑡𝑠

28
Page
𝑏) 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑒𝑜𝑝𝑙𝑒 𝑤ℎ𝑜 𝑒𝑛𝑡𝑒𝑟𝑒𝑑 200𝑚 = 33 (𝑠𝑢𝑚 𝑖𝑛 𝑡ℎ𝑒 𝑣𝑒𝑛𝑛 𝑑𝑖𝑎𝑔𝑟𝑎𝑚)
16
𝑐) 𝑃(𝐴𝑡ℎ𝑙𝑒𝑡𝑒 𝑟𝑢𝑛𝑠 2 𝑒𝑣𝑒𝑛𝑡𝑠) = (𝐴𝑙𝑙 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑠)
80

4. 𝑉𝑒𝑛𝑛 𝐷𝑖𝑎𝑔𝑟𝑎𝑚

8 42
𝑏) 𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒) = 1 − 𝑃(𝑟𝑒𝑎𝑑𝑠 𝑛𝑜𝑛𝑒) = 1 − = = 0.84
50 50
16 + 9 + 6
𝑐) 𝑃(𝑟𝑒𝑎𝑑𝑠 𝑜𝑛𝑙𝑦 𝑜𝑛𝑒) = 𝑃(𝐴 𝑜𝑛𝑙𝑦) + 𝑃(𝐵 𝑜𝑛𝑙𝑦) + 𝑃(𝐶 𝑜𝑛𝑙𝑦) =
50
31
𝑃(𝑟𝑒𝑎𝑑𝑠 𝑜𝑛𝑙𝑦 𝑜𝑛𝑒) = = 0.62
50
16
𝑑) 𝑃(𝑟𝑒𝑎𝑑𝑠 𝑜𝑛𝑙𝑦 𝐴) = = 0.32
50

1.11.4 Exercise 4

25 1
1𝑎) 𝑃(𝑙𝑎𝑡𝑒) = =
100 4
29

1𝑏) 𝑃(𝑜𝑛 𝑡𝑖𝑚𝑒) = 0.65

Page
 7
1𝑐) 𝑃(𝐴𝑏𝑠𝑒𝑛𝑡 𝑖𝑓 𝐵𝑜𝑦) =
10

25 5
1𝑑) 𝑃( 𝑜𝑛 𝑡𝑖𝑚𝑒 𝑖𝑓 𝑔𝑖𝑟𝑙) = =
65 13

25 10 35 7
1𝑒) 𝑃(𝐿𝑎𝑡𝑒 𝑜𝑓 𝐴𝑏𝑠𝑒𝑛𝑡) = + = =
100 100 100 20
25 7 95 19
1𝑓) 𝑃(𝐿𝑎𝑡𝑒 𝑜𝑟 𝐴𝑏𝑠𝑒𝑛𝑡 𝑖𝑓 𝐵𝑜𝑦) = + = =
100 10 100 20
305
2. 𝑎) 𝑃(𝑖𝑠 𝑎 𝑐𝑎𝑟 𝑝ℎ𝑜𝑛𝑒 𝑢𝑠𝑒𝑟) =
755

685
𝑏) 𝑃(𝑛𝑜 𝑠𝑝𝑒𝑒𝑑𝑖𝑛𝑔 𝑓𝑒𝑒𝑠) =
755
25
𝑐) 𝑃(𝑐𝑎𝑟 𝑝ℎ𝑜𝑛𝑒 𝑢𝑠𝑒𝑟 𝑖𝑓 𝑡ℎ𝑒 𝑝𝑒𝑟𝑠𝑜𝑛 ℎ𝑎𝑑 𝑠𝑝𝑒𝑒𝑑𝑖𝑛𝑔 𝑓𝑖𝑛𝑒) =
70

405
𝑑) 𝑃(𝑛𝑜 𝑠𝑝𝑒𝑒𝑑𝑖𝑛𝑔 𝑓𝑖𝑛𝑒 𝑔𝑖𝑣𝑒𝑛 𝑑𝑖𝑑 𝑛𝑜𝑡 𝑢𝑠𝑒 𝑐𝑎𝑟 𝑝ℎ𝑜𝑛𝑒) =
450

1.11.5 Exercise 5

1𝑎) 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑟𝑒𝑝𝑒𝑎𝑡𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑎𝑟𝑒 𝑡𝑎𝑘𝑒𝑛 𝑜𝑢𝑡
11!
𝑃𝑅𝑂𝐵𝐴𝐵𝐼𝐿𝐼𝑇𝑌 ℎ𝑎𝑠 11 𝑙𝑒𝑡𝑡𝑒𝑟𝑠 ℎ𝑒𝑛𝑐𝑒 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑙𝑒𝑡𝑡𝑒𝑟𝑠 𝐼 𝑎𝑛𝑑 𝐵 𝑎𝑟𝑒 𝑟𝑒𝑝𝑒𝑎𝑡𝑒𝑑
2! 2!
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑒𝑠 = 9 979 200

1𝑏) 𝐹𝑖𝑠𝑡 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑖𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑎𝑛𝑑 𝑙𝑎𝑠𝑡 𝑙𝑒𝑡𝑡𝑒𝑟𝑠 𝑎𝑟𝑒 𝐵:
𝑤𝑒 𝑛𝑜𝑤 ℎ𝑎𝑣𝑒 9 𝑙𝑒𝑡𝑡𝑒𝑟𝑠 𝑡𝑜 𝑟𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑒 𝑟𝑎𝑛𝑑𝑜𝑚𝑙𝑦:
9!
𝑃𝑅𝑂𝐴𝐼𝐿𝐼𝑇𝑌 ∶ 𝑠𝑖𝑛𝑐𝑒 𝐼 𝑖𝑠 𝑟𝑒𝑝𝑒𝑎𝑡𝑒𝑑 𝑎𝑛𝑑 𝑖𝑡 𝑔𝑖𝑣𝑒𝑠 181 440
2!
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 181 400 1
∴ 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒: =
9 979 200 55

2. 𝑎) 𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠 𝑓𝑜𝑟 𝑙𝑒𝑎𝑟𝑛𝑒𝑟𝑠 𝑡𝑜 𝑠𝑖𝑡 = 5! = 120

2𝑏) 𝐹𝑜𝑟 𝑡𝑤𝑜 𝑙𝑒𝑎𝑟𝑛𝑒𝑟𝑠 𝑡𝑜 𝑠𝑖𝑡 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟, 𝑡ℎ𝑒 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑎𝑟𝑒 2! × 4!
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑛𝑜𝑡 𝑠𝑖𝑡𝑡𝑖𝑛𝑔 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟 𝑖𝑠 5! − 2! × 4! = 72
72 3
2𝑐) 𝑃(𝑛𝑜𝑡 𝑠𝑖𝑡 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟) = =
120 5

3𝑎) 𝑃𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑤𝑖𝑡ℎ 𝑟𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑎𝑙𝑙𝑜𝑤𝑒𝑑 = 6 × 6 × 6 × 6 = 1296 𝑤𝑎𝑦𝑠

𝑏) 𝑅𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑛𝑜𝑡 𝑎𝑙𝑙𝑜𝑤𝑒𝑑 ≔ 6 × 5 × 4 × 3 = 360
𝑐) 𝑅𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑎𝑙𝑙𝑜𝑤𝑒𝑑 𝑏𝑢𝑡 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 2 = 6 × 6 × 6 × 3 = 648
𝑑) 𝑅𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑛𝑜𝑡 𝑎𝑙𝑙𝑜𝑤𝑒𝑑 𝑎𝑛𝑑 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 2 = 5 × 4 × 3 × 3 = 180

4𝑎) 𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠 𝑓𝑜𝑟 𝑀𝐴𝑇𝑅𝐼𝑋 = 6! = 720

4𝑏) 𝑙𝑎𝑠𝑡 𝑙𝑒𝑡𝑡𝑒𝑟 𝑖𝑠 𝑎 𝑣𝑜𝑤𝑒𝑙 = 2! × 5! 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠
2! × 5! 1
∴ 𝑃(𝑙𝑎𝑠𝑡 𝑙𝑒𝑡𝑡𝑒𝑟 𝑣𝑜𝑤𝑒𝑙) = =
6! 3
2! × 5! 1
4𝑐) 𝑇𝑤𝑜 𝑣𝑜𝑤𝑒𝑙𝑠 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟 = =
6! 3

5𝑎) 𝑊𝑎𝑦𝑠 𝑡𝑜 𝑔𝑟𝑜𝑢𝑝 𝐶𝐻𝐸𝑀𝐼𝑆𝑇𝑅𝑌 = 9! = 362 880

5𝑏) 𝑆𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 𝐶 𝑎𝑛𝑑 𝑒𝑛𝑑𝑖𝑛𝑔 𝑤𝑖𝑡ℎ 𝑌 = 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑎𝑟𝑟𝑎𝑛𝑔𝑒 𝐻𝐸𝑀𝐼𝑆𝑇𝑅
𝑛𝑢𝑚𝑏𝑒𝑟(𝐻𝐸𝑀𝐼𝑆𝑇𝑅) = 7! = 5040
5𝑐) 𝑀𝐼𝑆 𝑔𝑟𝑜𝑢𝑝𝑒𝑑 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟 = 3! × 7! = 30 240 31
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1.11.6 Exam Problems

1.1 𝑇𝑜 𝑠𝑒𝑒 𝑤ℎ𝑒𝑡ℎ𝑒𝑟 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡: 𝑃(𝐴) × 𝑃(𝐵) = 0.2 × 0.63
𝑃(𝐴) × 𝑃(𝐵) = 0.126 = 𝑃(𝐴 𝑎𝑛𝑑 𝐵) ℎ𝑒𝑛𝑐𝑒 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑒𝑣𝑒𝑛𝑡𝑠

1.2.1 𝑃𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑎𝑟𝑒 77 = 823 543 𝑤𝑎𝑦𝑠

1.2.2 𝐿𝑒𝑡𝑡𝑒𝑟𝑠 𝑛𝑜𝑡 𝑟𝑒𝑝𝑒𝑎𝑡𝑒𝑑 = 7! = 5040 𝑤𝑎𝑦𝑠
1.2.3 𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 3 𝑣𝑜𝑤𝑒𝑙𝑠 𝑖𝑛 𝐷𝐸𝐶𝐼𝑀𝐴𝐿, 4 𝑐𝑜𝑛𝑠𝑜𝑛𝑎𝑛𝑡𝑠 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 5
𝑙𝑒𝑡𝑡𝑒𝑟𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑒𝑑 𝑎𝑠 5! 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒: 3 × 5! × 4 = 1440

1.3 𝑊𝑒 𝑛𝑒𝑒𝑑 𝑎 𝑡𝑟𝑒𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑓𝑜𝑟 𝑖𝑡:

52
𝑃(𝑂𝑂) + 𝑃(𝑌𝑌) = 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔:
100
𝑡 𝑡 2 2 52
× + × = 𝑒𝑥𝑝𝑎𝑛𝑑𝑖𝑛𝑔 𝑎𝑛𝑑 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦𝑖𝑛𝑔 𝑔𝑖𝑣𝑒𝑠:
𝑡 + 2 𝑡 + 2 𝑡 + 2 𝑡 + 2 100
52(𝑡 2 + 4𝑡 + 4) = 100(𝑡 2 + 4) 𝑤ℎ𝑖𝑐ℎ 𝑟𝑒𝑑𝑢𝑐𝑒𝑠 𝑡𝑜
4
3𝑡 2 − 13𝑡 + 12 = 0 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑔𝑖𝑣𝑒𝑠 𝑡 = 3 𝑜𝑟 𝑡 = 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡 = 3.
3
32
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𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 3 𝑜𝑟𝑎𝑛𝑔𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑎𝑔.

2.1 𝑊𝑒 𝑛𝑒𝑒𝑑 𝑃( 𝐵): 𝑆𝑡𝑎𝑟𝑡 𝑤𝑖𝑡ℎ 𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) = 𝑃(𝐴) + 2𝑃(𝐴)
0.57
∴ 0.57 = 3𝑃(𝐴) 𝑎𝑛𝑑 𝑃(𝐴) = = 0.19
3

2.2.1 𝑇𝑟𝑒𝑒 𝐷𝑖𝑎𝑔𝑟𝑎𝑚

1 2 1
2.2.2 𝑃(𝐴𝑌) = × =
2 5 5
1 3 1 5 26
2.2.3 𝑃(𝑃) = × + × =
2 5 2 9 45
3.1 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑎 𝑎𝑛𝑑 𝑏: 𝑎 + 12 = 32 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 = 20
𝑏 = 80 + 48 = 128
20
3.2 𝑆𝑖𝑛𝑐𝑒 𝑃(𝑀 𝑎𝑛𝑑 𝑛𝑜𝑡 𝑤𝑎𝑡𝑐ℎ𝑖𝑛𝑔 𝑇𝑉) = ≠ 0 ℎ𝑒𝑛𝑐𝑒 𝑁𝑜 𝑖𝑡𝑠 𝑛𝑜𝑡
160
128 4
3.3.1 𝑃(𝑊𝑎𝑡𝑐ℎ𝑖𝑛𝑔 𝑡𝑣) = = = 0.8
160 5
12 3
33

3.3.2 𝑃(𝑓𝑒𝑚𝑎𝑙𝑒 𝑛𝑜𝑡 𝑤𝑎𝑡𝑐ℎ𝑖𝑛𝑔 𝑡𝑣) = = = 0.075

160 40
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4.1 𝑉𝑒𝑛𝑛 𝐷𝑖𝑎𝑔𝑟𝑎𝑚:

4.2 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥: ∑ 𝑒𝑣𝑒𝑟𝑦𝑡ℎ𝑖𝑛𝑔 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑒𝑡𝑠 = 100

49 − 𝑥 + 4 + 8 + 5 + 𝑥 + 60 − 𝑥 + 2 + 14 = 100 𝑎𝑛𝑑 𝑥 = 42

7 + 2 + 18 27
4.3 𝑃(𝑢𝑠𝑒 𝑜𝑛𝑙𝑦 𝑜𝑛𝑒 𝑎𝑝𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛) = = = 0.27
100 100

5.1 𝑇ℎ𝑒 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑎𝑟𝑒 ∶ 5 × 5 × 10 × 9 = 2250

5.2 𝑇𝑎𝑏𝑙𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛

34
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