CHAPTER III

 

PROBABILITY

 

1.0            Chance events:

 

In this chapter, we shall be concerned with chance events. For example, a weatherman makes a forecast of the future weather. His forecast, “Rain” is more accurately a probability statement.

 

“The Express Football team will win the Super division league”. But what you mean to say is, “It is likely that express will win the league.”

 

Probability has many practical uses. For example, the school uses probability in setting up budget requirements; Politicians use probability to work out their strategies in order to win elections.

 

In this section, we study some ideas about statements involving chance events, like “if I toss a coin and allow it to fall freely, it will show heads” or “Gayaza will win the Schools Lawn tennis Championships”.

We will concern ourselves with a measure of chance that an event will happen. This measure of chance is also called the probability that the event will occur.

 

Definition:

 

Probability is a measure of the likelihood of a required outcome happening. It is usually given as a fraction.

 

Probability = number of required outcomes

Number of possible outcomes.

 

Example 1

If a coin is tossed once, find the probability that a head appears.

 

Possible outcomes = { Head, tail }

Required outcomes = { Head }

Therefore probability that head appears, denoted as P(Head) = ¹/₂

 

Example 2: A die is rooled once (i) Find the probability that an even number appears.

Possible outcomes = {1, 2, 3, 4, 5, 6}

Required outcomes, “set of even numbers”

= { 2, 4, 6}

P(even numbers) = 3 = 1

6                  2

 

(ii) Find the probability that a prime number appears

required outcome, “set of prime numbers”

= { 2, 3, 5 }

P(Prime) = 3 = 1

6 2

 

(iii) Find the probability that an even prime number appears.

Set of even numbers = { 2 }

P(even prime number) = 1

6.

 

Example 3:

If there are n equally likely ways of doing a certain action and x if them produce a certain event A, then the theoretical probability of A happening is given by

P(A) = number of ways in which A happens

Number of ways in which the action is done.

= x

n.

The probability that event. A will not happen, denoted

P(A’) = n – x

N

Since there are n – x ways in which A does not happen,

Note: A’ means A does not happen and is read as A prime.

\ P(A’) = n – x

n

= 1 – x

n

but x = P(A)

n

\ P(A’) = 1 – P(A)

Þ                          P(A’) + P(A) = 1 ………………………………………..***

 

Which is a certainity since A must either happen or not happen.

If an event cannot possibly happen, then its probability = 0 (impossibility). For the probability of picking a white ball from a bag containing black balls only is 0. But the probability of picking a black ball is 1, 0 certainity.

 

Therefore if P is the probability of an event happening, then P lies in the range

0 £ P £ 1.

Example 4:

Sarah and Edith have played each other at tennis 15 times this season. Sarah has won 12 of the matches. They play each other in a championship. What is the probability that

(a)   the match is drawn?

(b)  Sarah wins?

(c)   Either Sarah or Edith wins?

 

Solution:

Tennis matches are either won or lost. They are never drawn.

(a)   P(draw) = 0

(b)  Sarah has won 12 of the last 15 matches

P(Sarah winning) = 12 = 4 = 0.8

15 5

This is an example of experimental probability. It is based on the number of matches won out of the total played.

 

(c)   Since one of either Sarah or Edith must win, the probability of either person winning = 1

 

Example 5:

A letter is chosen at random from the alphabet. Find the probability that it is: a) B

b)                 L or T

c)                  One of the letters of the word BETHEL

d)                 Not one of the letters of the word DIRECT

Note: “at random” means “in a free, irregular way”

 

1.2           MUTUALLY EXCLUSIVE EVENTS.

If the events A and B cannot happen at the same time, then they are said to be mutually exclusive events. That is to say an event (A) excludes the possibility of the other event (B) on the other way round.

Addition law.

If the events A, B, C, ……, are mutually exclusive, the probability of A or B or C or …. Happening is the sum of their individual probabilities.

 

P(A or B or C or …) = P(A) + P(B) + P(C) + …..

= P(A È B È C È …)

= P(A) + P(B) + P(C) – P(A Ç B) – P(B Ç C) – P(A Ç C) – P(A Ç B Ç C)

 

Example 1:

Find the probability that a letter chosen at random from the alphabet is either a vowel or one of the letters P, Q, R, S. Consider the 2 sets;

Vowels, V = { A, E, I, O, U} B = { P, Q, R, S}

 

\P(V or B) = P(V) + P(B) = 5 + 4 = 9

26 26 26.

P(V È B) = P(V) + P(B) – P(V Ç B); P(V Ç B) = 0

 

Note: The set of possible outcomes is also called the sample space.

 

Example:

What is the sample space of an expectant mother?

 

Sample space = { baby boy, baby girl }

(Possible outcomes)

 

1.3           Independent events.

If the happening of event A has no effect on the possible happening of another event B, then A and B are said to be independent events. In such cases, the separate probabilities are multiplied to give a Combined probability.

Product law.

If events A, B, C are independent, the probability of A and B and C …. Happening is the product of their individual probabilities.

 

P(A and B and C …) = P(A) X P(B) X P(C)

 

P(A Ç B Ç C) = P(A) X P(B) X P(C).

 

Example:

A bag contains 7 black balls and 5 white balls. A ball is drawn from the bag and replaced, then a second one is drawn. What is the probability that: a) one is black and one is white.

b)                 atleast one is black?

 

Solution. There are 12 balls of which 7 are black.

P(drawing a black) = 7

12.

Since the first ball is replaced, then at the second choice, there are 12 balls of which 5 are white.

P(grawing a white) = 5

12.

The colour drawn second is independent of the colour drawn first.

\Probability of drawing first a black ball and then a white ball

= 7 X 5 = 35

12 12 144

But these two cases are mutually exclusive events;

\ Probability of drawing a black ball and a white ball when the ball order does not matter

= 35 + 35 = 70 = 35

144 144 144 72

b)                 Probability of atleast one black ball

= 1 – probability of no black ball.

= 1 – 5 X 5

12        12

= 1 - 25

144

= 119

144.

Notice that the product law is used to solve problems, which contain the words and or both/and.

 

1.4           Outcome table, tree diagrams.

Example: Two dice are thrown at the same time. Find the probability of getting:

a)                 at least one 4

b)                 a total score which is prime.

c)                  Probability of getting a score of (i) 3

(ii) 5

(iii) 7

(iv) 11

The table below shows all the possible outcomes when two dice are thrown.

	1	2	3	4	5	6
1
2
3
4
5
6

Note: There are 36 possible outcomes.

For the one die, there are 6 outcomes.

For the two dice, there are 6² = 36 outcomes.

For three dice, there are 6³ outcomes

Also for a coin, ther are 2 outcomes

For two coins, 2² = 4 outcomes

 

	H	T
H
T

 

And for 3 coins, 2³ = 8 outcomes

 

In the experiments of tossing a coin and throwing a die, the symmetry of a coin tells us that “heads” and “tails” should come down about equal number of times.

P(head) = P(tail)

Also for a fair die, all the numbers should show up about equal number of times.

P(1) = P(2) = P(3) = P(4) = P(5) = P(6).

We say that these events are equally likely. Probability found by such an argument is called theoretical probability.

Also P(A È B ) = P(A) + P(B) – P(A Ç B), for any two events A and B.

 

TREE DIAGRAMS.

Example A bag contains 3 black balls (B) and 2 white balls (W).

a)                 A ball is taken from the bag and then replaced. A second ball is chosen. What is the probability that

(i)                 they are both blsck?

(ii)              One is black and one is white?

 

b)                 Find out how those probabilities are affected if two balls are chosen without any replacement.

 

(i)                 With replacement.1^st choice 2^nd choice

P(B)=³/₅ BB

 

P(B)= ³/₅

P(W) = ²/₅ BW

WB

P(B)=³/₅

P(W)=²/₅

 

P(W)= ²/₅ WW

 

 

(ii)              Without replacement.

1^st Choice 2^nd Choice

 

P(B) = ³/₅ P(B)=³/₄ BB

 

 

 

P(W)=²/₄ BW

 

P(W)=²/₅ P(B)= ³/₄ WB

 

 

 

 

P(W) = ¹/₄ WW

 

 

Then use the tree diagrams to answer the problems.

 

Questions on Probability.

 

1.                 If two dice a re thrown, find the probability of getting

a) an odd score b) a score more than 7

 

2. A bag contains 4 red mables and 5 blue marbles. If one marble is picked and replaced and then another marble is picked. What is the probability of a) both being red b) one of each colour.

 

3.When 4 coins are tossed, what is the probability of

a)     4 heads

b)     2 heads and 2 tails

c)      at least one head.

4. If the probability of one student solving a problem is 6/7, of a second student solving it is 7/8 and of the third student solving is 8/9. Find the probability that.

a)     they will solve it

b)     at least one will solve it.

 

5. The ratio of red to yellow to green balls in a large bag is 5:4:6. Two balls are picked at random.

Find the probability that a) they are both green

b) they are of different colours

6. If 3 cards are drawn from a pack without replacement, find the probability that, a) they are all red

b) at least one is red

7. A box of eggs contains 9 good ones and 3 cracked ones. Find the probability that out of 3 eggs from the box,

a) they are all good b) one is good and 2 are cracked.