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Probability is, in essence, the mathematics of chance. How sure can we be that a particular event is going to happen? When we go to bed at night, we can be sure that the sun will rise the next morning. If today is Monday, then we can be certain that tomorrow is Tuesday. The chance that these events will occur is 100%. The probability of these events is 1.

Some events could possibly happen, but it is unlikely. For example: The chance that you win the lotto if you have bought a ticket is very slim, but it could happen. The chance that a lion is seen in the middle of town is very low, but it could happen if the circus was in town. The probability of these events lies between 0 and 1. A probability of ½ means a 50-50 chance of it happening. Then you get some events that will never happen. For example: the chance of throwing a 10 on a normal die is 0%. A die only goes up to 6. The probability of this is 0.

Probability of a single event = Number of ways in which an event can occur/ Number of possible outcomes.

Example: Sipho’s house has 1 lounge, 1 bathroom, 1 kitchen and 2 bedrooms.

What is the probability that Sipho is in the kitchen?

Probability = Number of ways in which an event can occur divided by the Number of possible outcomes.

= 1 kitchen ÷ 5 room

= 0.2

What is the probability that Sipho is in a bedroom?

Probability = Number of ways in which an event can occur divided by the Number of possible outcomes.

= 2 bedrooms ÷ 5 rooms

= 0.4

Two or more occurrences happen independently of each other.

Sometimes two or more events happen at the same time. Let us take the example of Clive and his coloured socks again. This time, Clive is also trying to put on a T-shirt. He has the following items:

What is the probability of Clive putting on a yellow shirt and yellow socks?

We work out each probability separately and then multiply them to get the total probability.

P (yellow socks) = 2 ÷15

P (yellow T-shirt) = 3 ÷ 20

P (socks and shirt) = 2 ÷ 15 x 3 ÷ 20 = ^{1}/_{50 }

What is the probability that he will wear black socks and a red shirt?

P (black socks) = 4 ÷ 15

P (red T-shirt) = 6 ÷20

P (socks and shirt) = 4 ÷15 x 6 ÷ 20

= ^{2}/_{25}

*When two or more occurrences happen independently of each other, then we multiply the probabilities.*

Example: A drawer contains 3 red paperclips, 4 green paperclips, and 5 blue paperclips. One paperclip is taken from the drawer and then replaced. Another paperclip is taken from the drawer.

What is the probability that the first paperclip is red and the second paperclip is blue?

Because the first paper clip is replaced, the sample space of 12 paperclips does not change from the first event to the second event. The events are independent.

P (red then blue) = P (red) x P (blue)

= 3÷12 x 5÷12

= 15÷144

= ^{5}/_{48}

In a parking lot there are 20 vehicles. 16 of the vehicles are cars.

What is the probability of the first vehicle leaving the parking lot being a car?

P(car) = 16 ÷20

= ^{4}/_{5 }

What is the probability of the first vehicle leaving the parking lot NOT being a car?

P (Not car) = 4÷20

= ^{1}/_{5 }

Notice that: 4 ÷5 + 1÷5 = 5÷5 = 1

P(event) + P(not event) = 1

Either one or the other event happening.

Either one or the other event happening You have a deck of 52 playing cards. The jokers have been removed. A card is drawn from the deck.

What is the probability of the card being red or a Jack?

Quite clearly: P (red) = 13÷52 and P (Jack) = 4÷52

But, there are two red jacks in the deck!! You cannot count those twice.

Their probability is ^{2}/_{52}.

Thus we use the formula:

P (red or Jack) = P (red) + P (Jack) – P (red and Jack)

= 13÷52 + 4÷52 – 2÷52

= ^{15}/_{52 }

*If you need to calculate the probability of either one or another event, then you use the formula: P(X or Y) = P(X) + P(Y) – P(X and Y)*

Putting this all together:

Example: A pair of dice is rolled.

What is the probability that the sum of the numbers rolled is either 7 or 11?

Six outcomes have a sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) P(7) = ^{6}/_{36 }

Two outcomes have a sum of 11: (5,6), (6,5) P(11) = ^{2}/_{36 }

The sum of the numbers cannot be 7 and 11 at the same time, so these events are mutually exclusive.

P(7 or 11) = P(7) + P(11)

= 6÷36 + 2÷36

= 8÷36

= ^{2}/_{9 }

Example: A pair of dice is rolled. What is the probability that the sum of the numbers rolled is either an even number or a multiple of 3?

Of the 36 possible outcomes, 18 are even sums. P(even) = 18÷36

= ^{1}/_{2 }

Sums of 3, 6, 9, and 12 are multiples of 3. There are 12 sums that are multiples of 3.

P (multiple of 3)= 12÷36

= ^{1}/_{3 }

However, some of these outcomes appear in both events.

The sums that are even and a multiple of 3 are 6 and 12.

There are 6 ordered pairs with these sums.

P(even AND a multiple of 3) = 6÷36

= ^{1}/_{6 }

P(even OR a multiple of 3) = 18÷36 + 12÷36 - 6÷36

= 24÷36

= ^{2}/_{3}

Two events are said to be independent if the result of the second event is not affected by the result of the first event.

If A and B are independent events, the probability of both events occurring is the product of the probabilities of the individual events.

If A and B are independent events, then P(A and B) = P(A) x P(B).

Example: A drawer contains 3 red paperclips, 4 green paperclips, and 5 blue paperclips. One paperclip is taken from the drawer and then replaced. Another paperclip is taken from the drawer.

What is the probability that the first paperclip is red and the second paperclip is blue?

Because the first paper clip is replaced, the sample space of 12 paperclips does not change from the first event to the second event. The events are independent.

P(red then blue) = P(red) x P(blue)

= 3÷12 X 5÷12

= 15÷144

= ^{5}/_{48}

If the result of one event IS affected by the result of another event, the events are said to be dependent.

If A and B are dependent events, the probability of both events occurring is the product of the probability of the first event and the probability of the second event once the first event has occurred.

If A and B are dependent events, and A occurs first, P(A and B) = P(A) x P(B), once A has occurred).

Example: A drawer contains 3 red paperclips, 4 green paperclips, and 5 blue paperclips. One paperclip is taken from the drawer and is NOT replaced. Another paperclip is taken from the drawer.

What is the probability that the first paperclip is red and the second paperclip is blue?

Because the first paper clip is NOT replaced, the sample space of the second event is changed. The sample space of the first event is 12 paperclips, but the sample space of the second event is now 11 paperclips. The events are dependent.

P(red then blue) = P(red) x P(blue)

= 3÷12 x 5÷11

= 15÷132

= ^{5}/_{44}