Probability

Probability

Experimental probability =  No of required outcomes /  No of possible outcomes

Ex 1 A die is rolled 200times. The outcomes obtained are as shown below

find the probability of obtaining a

1. a) 2 b) 5 c) 6
2. a) Probability of obtaining 2 is

=  = O.15

1. b) Probability of obtaining a 5 is

=

1. c) Probability of a 6 is

=

Theoretical Probability

Theoretical probabilities are exact values which can be calculated by considering the physical nature of the given situation.

For Instance, the given situation for instance, the probability of getting a  when a fair six-sided die is thrown is 1/6. Since any one of the six faces is equally likely. This is an example of theoretical probability.

Ex 2: Jessie throws a fair six-sided die what is probability that she throws a1) 4 9 (b) a4 (c) a number greater than 2 (d) an even number (e) either 1,,2,3,4,5,or 6.

Solution

1. a) Since the faces of six- sided die are numbers 1-6

It is impossible to throw a 9

Probability +0

(b) Probability of a 4 = no of required outcome

no of possible outcome

= 1/6

(c) There are 4 numbers greater than 2

Prob. (a no greater than 2) = 4/6 = 2/3

(d) No of even numbers =3

P (an even number) = 3/6 = ½

(e) P(Either 1, 2,3,4,5 or 6) =

Note: If P is the probability of an event happening then p lies in the range o<

the probability of an event not happening is 1 – P.

Ex. 3. A letter is chosen at random from the alphabet. Find the probability that it is 9a) F (b) F or T (c) one of the letters of the word FREQUENCY (d) Not one of the letters of the word Table.

Solution

1. a) no. of f in alphabet = 1

p (F) = 1/26

1. b) P ( F or T) = p(f) +p (T)

+   =  =

(c) No.of letters in the word FREQUENCY = 8

p(one of letters in the word FREQUENCY=  =

1. d) no.. of letters in the word TABLE = 5

p(one of letters in the word TABLE) = 5/26

p(not one of the letters in the word TABLE) =1-5/56

=21/26

Note: ‘At random’ means in a free, irregular way

Exercises

1. A statical survey shows that 28% of all men take size 9 shoes. What is the probability that your friend’s father takes size 9 shoes?

Solution

prob (that a person takes size 9 shoes) =

2. A school contains 357 boys and 323 girls If a student is chosen at random, What is the probability that a girl is chon?

Solution

No. of girls =323

Total no. of students = 323 + 357 =680

p (a girl is chosen) = 323 =193

680    40

3. A letter is chosen at random fro m the alphabet. Find the probability that it is
4. a) M
5. b) not A or Z
6. c) either P,Q,R or S
7. d) one of the letters

Solution

n(alphabets) =26

1. a) no. of m in the alphabets = 1

P(M) = 1/26

P (A) =  1/26

P(A or Z) =

P( not A or Z) = 1

1. c) P (either P,Q,R or S) =
2. d) P (one of the letters of Nigeria) =