PROBABILITY: SAMPLE SPACE,EVENT SPACE, INDEPENDENT AND DEPENDENT EVENTS

Probability: Is a measure of the likelihood that an event will occur in any one trial. Probability can be applied in several areas like insurance, industrial quality control and so on.

It could also be defined as the ratio of required outcome to the total outcome.

Probability = No of required outcome / Total outcome

Sample Space: This is the set of all possible outcomes of any random experiment, and it’s denoted by S. The number of outcomes is denoted by n(S).

Event Space: This is the collection of outcomes of a random experiment. The number is denoted by n(E).

Outcome: This is result of an experiment in probability.

The probability that an event is certain to happen is 1, while the probability that an event is certain not to happen is zero (0).

Range of inequality: 0 <pr (E)<1

Hence; Prob (an event will occur) + prob (an event will not occur) = 1.

EXAMPLE:

1. From a box containing 2 red,6 white and 5 black balls, a ball is randomly selected . What is the Probability that the selected ball is (i) black (ii) white (iii) not black?

Solution:

Sample space = 2+ 6 + 5 = 13

n(red) = 2,       n(white) = 6,    n(black) = 5

• Prob. (black) = 5/13                        (ii) Prob. (white) = 6/13

(iii)       Prob. (not black) = 1 – prob. (black) (iv)                                        = 1 – 5/13 = 8/13

• What is the probability that an integer selected from the set of integers {20, 21, …, 30} is a prime number?

Evaluation: 1. A box contains five 10 ohm resistors and twelve 30 ohm resistor. The resistors are all unmarked and of the same physical size. If one resistor is picked out at random, determine the probability of its resistance being 10ohms.

EQUIPROBABLE SAMPLE SPACE:

Coin: A coin has two faces called the head (H) and the tail (T). The outcome of the experiment involving a coin depends on the numbers of trials.

In a single throw of a fair coin: {H, t} = 2

Two coins thrown once or a coin thrown twice: {HH, HT, TH, TT} = 4

Three coins of thrown thrice: {HHH, HHT, HTT, HTH, THH, THT, TTH, TTT} = 8. Example: In a single throw of two fair coins. Find the probability that : (i) two tails appear (ii) one head one tail appear (iii) two heads appear (iv) one tail one head in that order.

Solution:

S= {HH, HT, TH, TT} = 4

• Pr (two tails) = {TT} = 1/4
• Pr (=one head one tail) = {HT, TH} = 2/4 = 1/2 (iii)Pr (two heads) = {HH} = 1/4

(iv)Pr (1 head 1 tail in that order) = (HT) = 1/4

Die:

A fair die is a six faced die. A die could be tossed in a different number of trials. When a die is tossed once: {1, 2, 3, 4, 5, 6} = 6

When a die is tossed twice or 2 dice tossed, the outcome represented in the table below and the total outcome is 36.

EVALUATION:

1. Two fair dice are tossed. Find the probability (a) of not getting a total of 9 (iii) that the two dice show the same number.
2. In a single throw of three fair coins, find the probability that: (a) one head two tails appear (b) at least one head appears.

MUTUALLY EXCLUSIVE EVENTS: Two events are mutually exclusive if they cannot occur at the same time. That is no common element between them. This leads to the ADDITION RULE.

Addition rule: :If E1, E2, E3,….. are dependent events than Pr(E1 UE2…. UEn) = Pr (E1) + Pr (E2)

+ …. Pr(En)

Words such as; or, either are used to indicate addition of Probabilities.

Example: In a single throw of a fair die, what is the probability that an even number or a perfect square greater than 1 shows up?

Solution: S = {1, 2, 3, 4, 5, 6} = 6

Even number = {2, 4, 6} = 3, perfect square > 1 = {4} Pr (even nos) = 3/6                      Pr (Perfect square> 1) = 1/6

Pr (even of perfect square) = 3/6 + 1/6

= 2/3

INDEPENDENT AND DEPENDENT EVENTS

Independent event: Tow or more events are said to be independent when the occurrence of one event does not affect the occurrence of the other events in any way. Hence, the events can occur independently. E .g obtaining a 6 in a single throw of a die and obtaining a tail in an event of coin. This leads to the MULTIPLICATION RULE.

Multiplication rule:If E1, E2, E3,….. En are independent events then Pr(E1nE2…nEn)=Pr(E1) x

Pr(E2) x …Pr (En)

When two events are combined with words such as; AND, BOTH. The probabilities of the events are multiplied.

An event is said to independent when picking is done with replacement.

Dependent Event: Two of more events are dependent when the occurrence of event 1 affects the occurrence of the other event (s). It is dependent when it is done without replacement.

Conditional Probability: This is the application of the multiplication rule.

Example:

1. A class consists of 8 men and 7 ladies. Two students were selected randomly to represent the class in a debate. Find the probability that two students selected are (a) both ladies (b) both men (c) of the same sex (d) of different sexes.

Solution:

Total students = 8+ 7 =15

n(men) = 8, n(ladies) = 7

1. Pr(both ladies) = 7/15 x 6/14 =1/5

b) Pr(both men) = 8/15 x 7/14 =4/15

• Pr(same sex) = Pr(m & m) or Pr (L & L) 4/15 + 1/5 = 7/15
• Pr (different sexes) = P (m & L) or P (L or m)

= (8/15 x7/14) + (7/15 x 8/14)

= 4/15 + 4/15 = 8/15

• Two marbles are selected randomly, from a box containing 5 red, 7 white and 8 blue marbles one after the other without replacement. Find the Probability that: (a) both are red (b) one is white and the other is blue. (c0 both are of the same colour.

Solution:

Total Marbles = 5 + 7 + 8 = 20

N(R) = 5,         n(w) =7,          n(B) = 8

(a) Pr (both red) = P(RR) = 5/20 x 4/19 =1/19

• Pr (one white, one blue) = The arrangement is important. Pr(wB) or P(Bw)            = (7/20 x 8/19) + (8/20 x 7/19)

= 14/95 + 14/95 = 28/95

• Pr(same colour) = Pr(RR) or Pr(WW) or Pr (BB)

= (5/20 x 4/19) + (7/20 x 6/19) + (8/20 x 7/19)

= 20/380 + 42/380 + 56/380

= 118/380 =59/180

EVALUATION:

1. A box contains 2 white and 3 blue identical marbles. If two marbles are picked at random, one after the other, without replacement, what is the probability of picking two marbles of different colour?
2. A ball is picked at random from a bag containing 5 green balls, 3 white balls and 2 black balls. What is the probability that it is either green or black?

Evaluation: The probabilities that Bala and Uzoamaka will pass an examination are given as

0.8 and 0.9 respectively. Find the probability that: (a) both of them fail the examination (b0 at least one of them will pass this examination.

WEEKEND ASSIGNMENT

1. In a single throw of a fair coin ,find the probability that a head appears a) ¾ b) ½ c) 2/3 d)¼ Two fair dice are tossed , find the probability that the total score is
2. prime number a) 5/12 b) ¾  c) ½  d)  5/36 3) less than 6 a) 5/36 b) 5/24 c) 5/6 d) 5/18

A bag contains 6 red and 4 blue identical marbles, if two marbles are picked one after the other without replacement, find the probability that both marbles are of

4) the same colour a) 8/15 b) 7/15 c) 1/3 d) 4/15

5) differentcolour a) 4/15 b) 7/15 c) 8/15 d) 4/9

THEORY

1. Two  dice are thrown  together,  what  is  the probability of getting     (i) at least 6 (ii) score  greater than 8
2. A box contains 5 white and 3 black balls , if two balls are drawn one after the other with replacement what is the probability that both of them are (i) of the same colour (ii) of different colour

EVALUATION

1. In how may ways can 8 students be seated in a row?
2. In how many ways can the 1^st, 2^nd 3^rd prizes be won by 6 athletetes in a race? 3.In how many ways can the letters of the word HISTORY be arranged?

CYCLIC PERMUTATION: Cyclic permutation is the arrangement of things around a circular object. Since a circular table has no beginning and no end, the number of arrangement is 1 x (n – 1) !

CONDITIONAL PERMUTATION:

Sometimes restrictions are placed on the order of arrangements of objects

READING ASSIGNMENT

Read permutation and combination, further mathematics project 2 pages 47-54

WEEKEND ASSIGNMENT

1.         Evaluate ⁶C2 + ⁶C3 + ⁶C4 + ⁶C5 (a) ⁶C6 (b) ⁶C5 (c) ⁸C5

• How much ways can the letters of the word EVALUATE be arranged? (a) 10080 (b) 20160 (c) 40320
• In how many ways can 2 boys and 3 girls be arranged to sit in a row, if the boys must sit together (a) 6 (b) 4 (c) 24
• Find the number of ways 6 people can be seated in a round table, if two particular friends must sit next to each other (a) 48 9b) 24 (c) 120
• In how many ways can 6 pupils be lined up if 3 of them insist on following one another? (a) 720 (b) 144 (c) 24

THEORY

1. Out of 7 lawyers, 5 judges, a committee consisting of 3 lawyers, 2 judges is to be formed, in how many ways can this be done, if
   1. Any lawyer and any judge can be included
   1. One particular judge can be included
   1. Two particular lawyer cannot be in committee
2. If ⁿP3 / ⁿC2 = 6, find the value of n