Introduction

The likelihood of an occurrence of an event or the numerical measure of chance is called probability.

Experimental probability

This is where probability is determined by experience or experiment. What is done or observed is the experiment. Each toss is called a trial and the result of a trial is the outcome. The experimental probability of a result is given by (the number of favorable outcomes) / (the total number of trials)

Example

A boy had a fair die with faces marked 1to6 .He threw this die up 50 times and each time he recorded the number on the top face. The result of his experiment is shown
below.

What is the experimental provability of getting?
a.)1 b.) 6

Solution

a.) P(Event) =
P(1)= 11/50

b.) P(4)= 9/50

Example

From the past records, out of the ten matches a school football team has played, it has won seven.How many possible games might the school win in thirty matches ?.

Solution
P(winning in one math) = 7/10.
Therefore the number of possible wins in thirty matches = 7/10 x 30 = 21 matches.

Range of probability Measure

If P(A) is the probability of an event A happening and P(A’) is the probability of an event A not happening, Then P(A’)= 1 – P(A) and P(A’) + P(A)= 1

Probability are expressed as fractions, decimals or percentages.

Probability space

A list of all possible outcomes is probability space or sample space .The coin is such that the head or tail have equal chances of occurring. The events head or tail are said to be equally likely or eqiprobable.

Theoretical probability

This can be calculated without necessarily using any past experience or doing any experiment. The probability of an event happening #number of favorable outcomes /total number of outcomes.

Example

A basket contains 5 red balls, 4green balls and 3 blue balls. If a ball is picked at random from the basket, find:
a.)The probability of picking a blue ball
b.) The probability of not picking a red ball

Solution

a.)Total number of balls is 12
The number of blue balls is 3

b.) therefore, P (a blue ball) =3/12

c.)The number of balls which are not red is 7.
Therefore P ( not a red ball)= 7/12

Example

A bag contains 6 black balls and some brown ones. If a ball is picked at random the probability that it is black is 0.25.Find the number of brown balls.

Solution

Let the number of balls be x
Then the probability that a black ball is picked at random is 6/x
Therefore 6/x = 0.25
   x = 24
The total number of bald is 24
Then the number of brown balls is 24 – 6 =18

Note:

When all possible outcomes are count able, they are said to be discrete.

Types of probability

Combined Events : These are probability of two or more events occurring

Mutually Exclusive Events : Occurrence of one excludes the occurrence of the other or the occurrence of one event depend on the occurrence of the other.. If A and B are two mutually exclusive events, then ( A or B) = P (A) + P (B).
For example when a coin is tossed the result will either be a head or a tail.

Example
i.) If a coin is tossed ;
P(head) + P( tail)
=

Note:

If [OR] is used then we add

Independent Events: Two events A and B are independent if the occurrence of A does not influence the occurrence of B and vice versa. If A and B are two independent events, the probability of them occurring together is the product of their individual probabilities .That is;

P (A and B) = P (A) x P(B)

Note;
When we use [AND] we multiply ,this is the multiplication law of probability.

Example
A coin is tosses twice. What is the probability of getting a tail in both tosses?

Solution
The outcome of the 2nd toss is independ of the outcome of the first .
Therefore;
P (T and T ) = P( T) X P( T)
= =

Example

A boy throws fair coin and a regular tetrahedron with its four faces marked 1,2,3 and 4.Find the probability that he gets a 3 on the tetrahedron and a head on the coin.

Solution
These are independent events.
P (H) = P(3) =

Therefore;
P (H and 3) = P (H) x P (3)
= ½ x ¼
= 1/8

Example
A bag contains 8 black balls and 5 white ones.If two balls are drawn from the bag, one at a time,find the probability of drawing a black ball and a white ball.
a.) Without replacement
b.) With replacement

Solution
a.) There are only two ways we can get a black and a white ball: either drawing a white then a black,or drawing a black then a white.We need to find the two probabilities;
P( W followed by B) = P (W and B)
=
b.) P(B followed by W) = P (B and W)

Note;
The two events are mutually exclusive, therefore.
P (W followed by B) or ( B followed by W )= P( W followed by B ) + P ( B followed by W)
= P (W and B) + P( B and W)
=
Since we are replacing, the number of balls remains 13.
Therefore;
P (W and B) =
P ( B and W) =

Therefore;

P [(W and B) or (B and W)] = P (W and B) + P (B and W)
=

Example
Kamau ,Njoroge and Kariuki are practicing archery .The probability of Kamau hitting the target is 2/5,that of Njoroge hitting the target is ¼ and that of Kariuki hitting the target is 3/7 ,Find the probability that in one attempt;
a.) Only one hits the target
b.) All three hit the target
c.) None of them hits the target
d.) Two hit the target
e.) At least one hits the target

Solution
a.) P(only one hits the target)
=P (only Kamau hits and other two miss) =2/5 x 3/5 x 4/7
= 6/35
P (only Njoroge hits and other two miss) = 1/4 x 3/5 x 4/7
= 3/35
P (only Kariuki hits and other two miss) = 3/7 x 3/5 x ¾
= 27/140
P (only one hits) = P (Kamau hits or Njoroge hits or Kariuki hits)
= 6/35 + 3/35 +27/140
= 9/20

b.) P ( all three hit) = 2/5 x 1/4 x 3/7
= 3/70

c.) P ( none hits) = 3/5 x 3/4 x 4/7
= 9/35

d.) P ( two hit the target ) is the probability of ;
Kamau and Njoroge hit the target and Kariuki misses = 2/5 x 3/7 x 4/7
Njoroge and Kariuki hit the target and Kamau misses = 1/4 x 3/7 x 3/5
Or
Kamau and Kariuki hit the target and Njoroge misses = 2/5 x 3/7 x 3/4
Therefore P (two hit target) = (2/5 x 1/4 x 4/7) + (1/4 x 3/7 x 3/5) + (2/5 x 3/7 x 3/4)
= 8/140 + 9/140 + 18/140
= ¼

e.) P (at least one hits the target) = 1 – P ( none hits the target)

= 1 – 9/35
= 26/35
Or
P (at least one hits the target) = 1 – P (none hits the target)
= 26/35

Note;
P (one hits the target) is different from P (at least one hits the target)

Tree diagram

Tree diagrams allows us to see all the possible outcomes of an event and calculate their probality.Each branch in a tree diagram represents a possible outcome .A tree diagram which represent a coin being tossed three times look like this;

From the tree diagram, we can see that there are eight possible outcomes. To find out the probability of a particular outcome, we need to look at all the available paths (set of branches).
The sum of the probabilities for any set of branches is always 1.

Also note that in a tree diagram to find a probability of an outcome we multiply along the branches and add vertically.
The probability of three heads is:

P (H H H) = 1/2 × 1/2 × 1/2 = 1/8
P (2 Heads and a Tail) = P (H H T) + P (H T H) + P (T H H)
= 1/2 × 1/2 × 1/2 + 1/2 × 1/2 × 1/2 + 1/2 × 1/2 × 1/2
= 1/8 + 1/8 + 1/8
= 3/8

Example
Bag A contains three red marbles and four blue marbles.Bag B contains 5 red marbles and three blue marbles.A marble is taken from each bag in turn.

a.) What is the probability of getting a blue bead followed by a red
b.) What is the probability of getting a bead of each color

Solution
a.) Multiply the probabilities together
P( blue and red) =4/7 x 5/8 = 20/56
=5/14
b.) P(blue and red or red and blue) = P( blue and red ) + P (red and blue)

= 4/7 x 5/8 + 3/7 x 3/8
= 20/56 + 9/56
=29/56

Example
The probability that Omweri goes to Nakuru is ¼ .If he goes to Nakuru, the probability that he will see flamingo is ½ .If he does not go to Nakuru, the probability that he will see flamingo is 1/3 .Find the probability that;
a.) Omweri will go to Nakuru and see a flamingo.
b.) Omweri will not go to Nakuru yet he will see a flamingo
c.) Omweri will see a flamingo

Solution
Let N stand for going to Nakuru ,N’ stand for not going to Nakuru, F stand for seeing a flamingo and F‘stand for not seeing a flamingo.

a.) P (He goes to Nakuru and sees a flamingo) = P(N and F)
= P(N) X P(F)
= ¼ X ½
= 1/8

b.) P( He does not go to Nakuru and yet sees a flamingo) =P( N’) X P( F)
= P (N’ and F)
= 3/4 X 1/3
= ¼
c.) P ( He sees a flamingo) = P(N and F) or P ( N’ and F)
= P (N and F) + P (N’ and F)
= 1/8 + 1/4
= 3/8

Past KCSE Questions on the Topic

1. The probabilities that a husband and wife will be alive 25 years from now are 0.7 and 0.9 respectively.
Find the probability that in 25 years time,
(a) Both will be alive
(b) Neither will be alive
(c) One will be alive
(d) At least one will be alive

2. A bag contains blue, green and red pens of the same type in the ratio 8:2:5 respectively. A pen is picked at random without replacement and its colour noted

(a) Determine the probability that the first pen picked is
(i) Blue
(ii) Either green or red

(b) Using a tree diagram, determine the probability that
(i) The first two pens picked are both green
(ii) Only one of the first two pens picked is red.

3. A science club is made up of boys and girls. The club has 3 officials. Using a tree diagram or otherwise find the probability that:
(a) The club officials are all boys
(b) Two of the officials are girls

4. Two baskets A and B each contain a mixture of oranges and limes, all of the same size. Basket A contains 26 oranges and 13 limes. Basket B contains 18 oranges and 15 limes. A child selected a basket at random and picked a fruit at a random from it.
(a) Illustrate this information by a probabilities tree diagram
(b) Find the probability that the fruit picked was an orange.

5. Two bags A and B contain identical balls except for the colours. Bag A contains 4 red balls and 2 yellow balls. Bag B contains 2 red balls and 3 yellow balls.

(a) If a ball is drawn at random from each bag, find the probability that both balls are of the same colour.

(b) If two balls are drawn at random from each bag, one at a time without replacement, find the probability that:
(i) The two balls drawn from bag A or bag B are red
(ii) All the four balls drawn are red