Probability

Probability

Probability

In our daily life, we often come across events which are concerned with the idea of the likelihood or the chance of occurrence of future events.

Consider the following events

A.     Amit is a Class X student in a certain school. There is a test in Mathematics on the coming Monday. We may ask "Will Amit pass the test?" Before we answer this question, we have to consider some past records. If Amit failed in all his past Mathematics tests, we say that he has a very slim chance of passing the test. But, if Amit passed all his past Mathematics tests, we say that he has a very great chance of passing the test.

B.    Prakash is a student of standard X in a certain school. If a student is chosen at random in standard X, will Prakash be chosen? Before we answer this question, we have to know about the present statistics, the number of students in standard X. If there are 45 students in the class, then Prakash's chance of being chosen is 1/45.

C.    If we toss a fair coin, is it more likely for a 'head' or a 'tail' to come up ? In this event, we can answer the question immediately without considering any past or present statistics. We all know that there is an equal chance for a 'head' or a 'tail' to come up. Thus the chance for a 'head' (or a 'tail') to come up is 1/2. An alternative word used for 'chance' is 'probability' and it is generally represented by "P".

Random Experiments

In most branches of knowledge, experiments are the ways of life. In probability and statistics, too, we concern ourselves with special type of experiments. We call these as random experiments.

A random experiment is an experiment in which

(i)     all the possible outcomes of the experiment are known in advance, and

(ii)    the exact outcome of any specific performance of the experiment is unpredictable, i.e., depends upon chance.

For example, Suppose a fair coin is tossed. There are two possible outcomes of the experiment; heads and tails. For any performance of the experiment one does not know what the exact outcome will be. Therefore, the tossing of a fair coin is a random experiment.

Sample space

The set of all possible outcomes of a random experiment is called the sample space. If x  S, then x is called as a sample point.

For instance, in illustration, the sample space is S₁ = {H, T} where H stands for heads and T for tails. If the same coin is tossed twice, then S₂ = {HH, HT,.TH, TT}. In case the coin is tossed three times, the sample space will be S₃ = {HHH, HHT, HTH, HIT, THH, THT, TTH, TTT). In case of a single toss, the sample space consists of 2 points. In case of two tosses, it consists of 2 x 2 = 2² = 4 points whereas in case of three tosses, it consists of 2 x 2 x 2 = 2³ = 8 points. In case of n tosses, the sample space contains 2ⁿ points.

To get all the points in the sample space S_n, corresponding to n tosses of a fair coin we first write S₁ = {H, T} corresponding to a single toss. To obtain S₂ we post fix H to each element in S₁ to obtain HH, TH and then postfix T to each element in S₁ to obtain HT, TT. These four elements constitute S₂. That is, S2 = {HH, TH, HT, TT}. Now, to obtain S₃ we postfix H to each element in S₂ to obtain HHH, THH, HTH, TTH and then T to each element in S₂ to obtain HHT, THT, HTT, TTT. These eight elements constitute S₃. That is, S₃ = {HHH, THH, HTH, TTH, HHT, THT, HTT, TTT}. In a similar way we can obtain S₄, S₅  etc. Alternatively we may use a tree diagram as shown below.

Thus, S₂ = {HH, HT, TH, TT}

and S₃ = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

Event

Any subset of a sample space is called an event. A sample space S serves as the universal set for all questions related to an experiment 'S and an event A w.r.t it is a set of all possible outcomes favorable to the event  A.

For example,

A random experiment: flipping a coin twice Sample space: S = {(HH), (HT), (TH), (TT)} The question: "both the flips show same face" Therefore, the event A: { (HH), (TT) } . See figure

Elementary and compound events

Single element subsets of sample space associated with a random experiment are known as elementary events.

Those subsets of sample space S associated to an experiment which are disjoint union of single element subsets of sample space S are known as compound events.

e.g.: Consider the experiment of throwing a die. The sample space associated with this experiment is S = {1, 2, 3, 4, 5, 6}. Events E₁ = {1}, E2 = {2}, E₃ = {3}, E₄ = {4}, E₅ = {5}, E₆ = {6} are elementary events whereas A_l = {2, 4, 6),

A₂ = {1, 3, 5}, A₃ = {3, 6) etc. are compound events.

Equally Likely Events

All possible results of a random experiment are called equally likely outcomes and we have no reason to expect any one rather than the other. For example, as the result of drawing a card from a well shuffled pack, any card may appear in draw, so that the 52 cards become 52 different events which are equally likely.

Mutually Exclusive Events

Events are called mutually exclusive or disjoint or incompatible if the occurrence of one of them precludes the occurrence of all the others. For example in tossing a coin, there are two mutually exclusive events viz turning up a head and turning up of a tail. Since both these events cannot happen simultaneously. But note that events are compatible if it is possible for them to happen simultaneously. For instance in rolling of two dice, the cases of the face marked 5 appearing on one dice and face 5 appearing on the other, are compatible.

Exhaustive Events

Events are exhaustive when they include all the possibilities associated with the same trial. In throwing a coin, the turning up of head and of a tail are exhaustive events assuming of course that the coin cannot rest on its edge.

Independent Events

Two events are said to be independent if the occurrence of any event does not affect the occurrence of the other event. For example in tossing of a coin, the events corresponding to the two successive tosses of it are independent. The flip of one penny does not affect in any way the flip of a nickel.

Dependent Events

If the occurrence or non-occurrence of any event affects the happening of the other, then the events are said to be dependent events. For example, in drawing a card from a pack of cards, let the event A be the occurrence of a king in the 1st draw and B be the occurrence of a king in the second draw. If the card drawn at the first trial is not replaced then events A and B are independent events.

Note:

(1)  If an event contains a single sample point i.e. it is a singleton set, then this event is called an elementary or a simple event.

(2)  An event corresponding to the empty set is an 'impossible event:

(3)  An event corresponding to the entire sample space is called a 'certain event'.

Complement of an event

The complement of an event A with respect to S is the set of all elements of S that are not in A. We denote the complement of A by the symbol A'. In the following figure, the shaded portion of S represents A' or A . We also call A' as "not A".

In the application of this definition, the term favourable is used rather loosely — favourable may mean that a patient may have a viral fever or a brand new television does not work or a book published by TMH company has more than 120 printing errors.

Thus probability is a concept which measures numerically the degree of certainty or uncertainty of the occurrence of an event.

For example, the probability of randomly drawing king from a well-shuffled deck of cards is 4/52. Since 4 is the number of favorable outcomes (i.e. 4 kings of diamond, spade, club and heart) and 52 is the number of total outcomes (the number of cards in a deck).

If A is any event of sample space having probability P, then clearly, P is a positive number (expressed as a fraction or usually as a decimal) not greater than unity.

ii) Similarly, the probability that the ball drawn is white

= P(W) = 4/15.

(iii)   The probability that the ball drawn is black = P(B) = 5/15 = 1/3.

(iv)   The probability that the ball drawn is not red

=1—P(R)=1-2/5 = 3/5

(v)    The probability that the ball drawn is red or white

                                    

E3. What is the chance that a leap year selected at random will contain 53 Sundays?

Sol. A leap year has 52 weeks and 2 more days.

The two days can be: Monday — Tuesday

Tuesday — Wednesday

 Wednesday — Thursday

Thursday — Friday

Friday — Saturday

Saturday — Sunday and

Sunday — Monday.

There are 7 outcomes and 2 are favorable to the 53rd Sunday. Now for 53 Sundays in a leap year, P(A) = 2/7.

Compound Probability

Two or more events are said to be independent if the occurrence or non-occurrence of any one of them does not affect the probabilities of occurrence of any of the others. Thus if a coin is tossed four times and it turns up a head each time, the fifth toss may be head or tail and is not influenced by the previous tosses. The probability that two or more independent events will happen is equal to the product of their individual probabilities. This is called the compound probability. Thus the probability of getting a head on both the fifth and sixth tosses is  1/2 x 1/2 = 1/4

Two or more events are said to be dependent if the occurrence or non-occurrence of one of the events affects the probability of occurrence of any of the others.

Consider that two or more events are dependent. If P₁ is the probability of a first event, P₂ the probability that after the first has happened the second will occur, P₃ the probability that after the first and second have happened the third will occur etc., then

the Probability that all events will happen in the given order is the product P₁ x P₂ x P₃ ....

For example, a box contains 3 white balls and 2 black balls. If a ball is drawn at random, the probability that it is black is 2/2+3 =  2/5  If this ball is not replaced and a second ball is drawn, the probability that it is also black is 1/3+1 = 1/4 . Thus the probability that both will be black is 2/5 x 1/4  = 1/10

Mutually Exclusive Events

Two or more events are said to be mutually exclusive if the occurrence of any one of them guarantees the non occurrence of the others.

The probability of occurrence of one or two or more mutually exclusive events is the sum of the probabilities of the individual events.

1.      In rolling a die

E:    The event that the number is odd.

F: The event that the number is even.

G:   The event that the number is a multiple of three.

2. In drawing a card from a deck of 52 cards

E:    The event that it is a spade.

F:    The event that it is a club.

G:   The event that it is a king.

In the above 2 cases, events E & F are mutually exclusive but the events E & G or F & G are not mutually exclusive or disjoint since they may have common outcomes.

Addition Law of Probability

If E & F are two mutually exclusive events, then the probability that either event E or event F will occur in a single trial is given by

ü  P(E or F) = P(E) + P(F)

If the events are not mutually exclusive, then

ü  P(E or F) = P(E) + P(F) — P(E & F together).

Note: Compare this with n(A  B) = n(A) + n(B) — n(A  B) of set theory.

Similarly

ü  P (neither E nor F) = 1— P(E or F).

    E4. A single card is selected from a deck of 52 bridge cards. What is the probability that

(1)       it is not a heart?

(2)       it is an ace or a spade?

Sol. A deck of bridge cards has 4 suits — spade, heart, diamond and club.

Each suit has 13 cards.— Ace, two, three, .... , ten, jack, Queen, King.

(1)       P (not a heart) = 1 — P (a heart) = 1 — 13/52 = 39/52 = 3/4.

(2)       There are 4 aces and 12 spades besides the ace of spade

P (an ace or a spade) = 16/52 = 4/13.

E5. If a die is thrown, what is the probability of getting a 5 or a 6?

Sol. Getting a 5 and getting a 6 are mutually exclusive.

So, P(5 or 6)=P(5)+P(6) =    1/6 + 1/6 = 2/6  =  1/3.

Two events are said to be overlapping if the events have at least one outcome in common, i.e., they can happen at the same time.

The probability of occurrence of only one of two overlapping events is the sum of the probabilities of the two individual events minus the probability of their common outcomes.

E6. If a die is thrown, what is the probability of getting a number less than 4 or an even number?

Sol. The numbers less than 4 on a die are 1, 2, and 3.

The even numbers on a die are 2, 4, and 6. Since these two events have a common outcome, 2, they are overlapping events.

P (less than 4 or even) = P(less than 4) + P(even)— P(less than 4 and even )

= 3/6 + 3/6 – 1/6 = 5/6.

Independent Events

Two events are independent if the happening of one has no effect on the happening of the other. For example:

1. On rolling a die & tossing a coin together

E:      The event that number 6 turns up.

F:      The event that head turns up.

2.     In shooting a target

E:      Event that the first trial is missed.

F:      Event that the second trial is missed. In both these cases events E & F are independent. But

3. In drawing a card from a well shuffled pack

E:      Event that first card is drawn

F:      Event that second card is drawn without replacing the first

G:      Event that second card is drawn after replacing the first In this case E & F are not independent but E & G are independent.

Multiplication Law of Probability

If the events E & F are independent then

ü  P(E & F together) = P (E) x P (F)

               & P (not E & F) = 1 — P(E & F together)

Odds In Favour And Against

If P is the probability that an event will occur and q (= 1— p) is the probability of the non-occurrence of the event; then we say that the odds in favour of the event occurring are p : q (p to q) and the odds against its occurring are q : p.

For example, if the event consists of drawing a card of club from the deck of 52 cards, then the odds in favour are

13/52 : 39/52 = ¼ : ¾ = 1:3

Similarly the odds against a card of diamond would be 3 : 1. The odds in favour of 4 or 6 in a single toss of a fair die are 1/3 : 2/3 i.e. 1: 2 and the odds against are 2 : 1.

Important

Ø It should be observed that the probability of the event B is increased due to the additional information that the event A has occurred.

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

E7.  One purse contains 5 coins of 50 paise and 2 coins of 25 paise, and a second purse contains 1 coin of 50 paise and 3 coins of 25 paise. If a coin is taken from one of the two purses at random, what is the probability that it is a coin of 25 paise?       

Sol. The probability of selecting the first purse (1/2) and then drawing a coin of 25 paise from it (2/7) is (1/2) (2/7) = 1/7.

The probability of selecting the second purse (1/2) and of then drawing a coin of 25 paise from it (3/4) is (1/2) (3/4) = 3/8.

Hence the required probability  = 1/7 + 3/8 = 29/56

E8.  A box contains black chips and red chips. A person draws two chips without replacement. If the probability of selecting a black chip and a red chip is 15/56 and the probability of drawing a black chip on the first draw is 3/4, what is the probability of drawing a red chip on the second draw, if you know the first chip drawn was black?

Sol. If B is the event drawing a black chip drawing a red chip, then P(R/B) is the probability of drawing a red chip on the second draw given that a black chip was drawn on the first draw.

Thus, the probability of drawing a red chip on the second draw given that a black chip was drawn on the first draw is  5/14.

E9. Two boxes A and B contain red and black balls. Box A contains 4 Red and 5 Black balls, while Box B contains 2 Red and 3 Black balls. One ball is randomly drawn from any of the two boxes. Find the probability of the ball being drawn from Box A if it is known that it is a Red ball.

Sol. Using Bayes Theorem, we can find the probability of a red ball being drawn from Box A = 4/9 and that of being drawn from Box B = 2/5. Thus the probability of the ball coming from Box A given it is a Red ball is

 =10/19

E10. Each of the two Amit and Rohit make statements - the probability of which being true is 1/3 each. Amit makes a statement and then Rohit says "Whatever Amit has said is the truth." What is the probability that Amit has actually spoken the truth.

Sol. Case I - If Amit makes a true statement (probability 1/3), then Rohit's statement is also true. 1/3 x 1/3 =1/ 9 .

Case II - If Amit makes a false statement (probability 2/3), then Rohit's statement is also false. 2/3 x 2/3 = 4/9  .

Using Bayes Theorem, we can find the probability of Amit actually making a true statement = 1/5

E11.Three urns contain 6 red, 4 black; 4 red, 6 black and 5 red, 5 black balls respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is red, find the probability that it is drawn from the first urn.

Sol. Let E₁, E₂, E₃ and A be the events defined as follows: E₁ = urn first is chosen, E₂ =urn second is chosen, E₃ = urn third is chosen, A = ball drawn is red.

Since there are three urns and one of the three urns is chosen at random, therefore P(E₁) = P(E₂) = P(E₃) = 1/3. If E₁ has already occurred, then urn first has been chosen which contains 6 red and 4 black balls. The probability of drawing a red ball from it is 6/10.

P(A/E₁) = 6/10 . similarly  , P (A/E₂) = 4/10 and P (A/E₃) = 5/10.

We are required to find P ( E ₁ / A ) , i.e., given that the ball drawn is red, what is the probability that it is drawn from the first urn. By Baye's rule

Try solving yourself through given formula

A bunch of few questions to solve for the crazy toppers

E12.In a bolt factory, machines A, B and C manufacture 25%, 35% and 40% of the total bolts respectively. Of their output 5, 4 and 2 per cent respectively are defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found to be defective, what is the probability that it is manufactured by the machine B?

 E13. A bag contains 5 white and 3 black balls. Four balls are successively drawn out without replacement. What is the probability that they are alternately of different colours ?

E14.A consignment of 15 record players contains 4 defective. The record players are selected at random, one by one, and examined. The ones examined are not put back. What is the probability that 9^th one examined is the last defective?

E15.Three persons A, B, C throw a die in succession till one gets a 'six' and wins the game. Find their respective probabilities of winning if A begins.