# Probability Concepts & Solved Examples For CAT - Nitin Gupta, AlphaNumeric (Part 1/2)

• In our day to day life, we perform many activities which have a fixed result no matter any number of times they are repeated. For example given any triangle, without knowing the three angles, we can definitely say that the sum of measure of angles is 180°.

We also perform many experimental activities, where the result may not be same, when they are repeated under identical conditions. For example, when a coin is tossed it may turn up a head or a tail, but we are not sure which one of these results will actually be obtained.

Random Experiment

An experiment is called random experiment if it satisfies the following two conditions:
(i) It has more than one possible outcome.
(ii) It is not possible to predict the outcome in advance.

Outcomes & Sample Space

A Possible result of a random experiment is called its outcome.

Consider the experiment of rolling a die. The outcomes of this experiment are 1, 2, 3, 4, 5, or 6, if we are interested in the number of dots on the upper face of the die.

The set of outcomes {1, 2, 3, 4, 5, 6} is called the sample space of the experiment.

Thus, the set of all possible outcomes of a random experiment is called the sample space associated with the experiment. Sample space is denoted by the symbol S.

Each element of the sample space is called a sample point. In other words, each outcome of the random experiment is also called sample point.

Two coins (a one rupee coin and a two rupee coin) are tossed once. Find a sample space.

Clearly the coins are distinguishable in the sense that we can speak of the first coin and the second coin. Since either coin can turn up Head (H) or Tail(T), the possible outcomes may be
Heads on both coins = (H,H) = HH
Head on first coin and Tail on the other = (H,T) = HT
Tail on first coin and Head on the other = (T,H) = TH
Tail on both coins = (T,T) = TT
Thus, the sample space is S = {HH, HT, TH, TT}

ORDERED & UN-ORDERED PAIRS

ORDERED PAIR : In case of ordered pairs HT,TH are different..
UN - ORDERED PAIRS : In case of un ordered HT n TH will be considered as 1 pair...

Basically we find unordered when the variables in the question are implicit.....In algebra "Ways" basically refers to finding UNORDERED n "Solutions" refers to finding ORDERED.

Because in " ways" (1,100) n (100,1) is same but in case of "Solution" (1,100) n (100,1) are different solutions....We ll go through the details when we explore algebra part....in questions like find the number of integral solutions to the equation 1/a + 1/b = 1/k.....we ll see this in algebra....i just thought that you guys should know the concept of "ordered n unordered" so told it here

Find the sample space associated with the experiment of rolling a pair of dice (one is blue and the other red) once. Also, find the number of elements of this sample space.

Suppose 1 appears on blue die and 2 on the red die. We denote this outcome by an ordered pair (1,2). Similarly, if ‘3’ appears on blue die and ‘5’ on red, the outcome is denoted by the ordered pair (3,5).

In general each outcome can be denoted by the ordered pair (x, y), where x is the number appeared on the blue die and y is the number appeared on the red die.

Therefore, this sample space is given by
S = {(x, y): x is the number on the blue die and y is the number on the red die}.
The number of elements of this sample space is 6 × 6 = 36 and the sample space is given below:
{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

In each of the following experiments specify appropriate sample space
(i) A boy has a 1 rupee coin, a 2 rupee coin and a 5 rupee coin in his pocket. He takes out two coins out of his pocket, one after the other.
(ii) A person is noting down the number of accidents along a busy highway during a year.

(i) Let Q denote a 1 rupee coin, H denotes a 2 rupee coin and R denotes a 5 rupee coin. The first coin he takes out of his pocket may be any one of the three coins Q, H or R. Corresponding to Q, the second draw may be H or R. So the result of two draws may be QH or QR. Similarly, corresponding to H, the second draw may be Q or R.
Therefore, the outcomes may be HQ or HR. Lastly, corresponding to R, the second draw may be H or Q.
So, the outcomes may be RH or RQ.
Thus, the sample space is S={QH, QR, HQ, HR, RH, RQ}
(ii) The number of accidents along a busy highway during the year of observation can be either 0 (for no accident ) or 1 or 2, or some other positive integer.
Thus, a sample space associated with this experiment is S= {0,1,2,...}

A coin is tossed. If it shows head, we draw a ball from a bag consisting of 3 blue and 4 white balls; if it shows tail we throw a die. Describe the sample space of this experiment.

Let us denote blue balls by B1, B2, B3 and the white balls by W1, W2, W3, W4. Then a sample space of the experiment is
S = { HB1, HB2, HB3, HW1, HW2, HW3, HW4, T1, T2, T3, T4, T5, T6}.
Here HBi means head on the coin and ball Bi is drawn, HWi means head on the coin and ball Wi is drawn. Similarly, Ti means tail on the coin.

NOTE:

Remember one thing --- in probability it doesn't matter whether the balls are identical or distinct
in p&c : if I ask you in how many ways you can pick 1 ball out of 3 identical balls -- there is exactly 1 way. but in probability. there are 3c1 = 3 ways
in probability --- we always look for no. of distinct ways

Consider the experiment in which a coin is tossed repeatedly until a head comes up. Describe the sample space.

In the experiment head may come up on the first toss, or the 2nd toss, or the 3rd toss and so on till head is obtained. Hence, the desired sample space is S= {H, TH, TTH, TTTH, TTTTH,...}

Event

The appearance of a particular outcome, which we may find in the sample space, is an event. Any subset of the sample space is called an event.

Occurrence of an event: Consider the experiment of throwing a die. Let E denotes the event “ a number less than 4 appears”. If actually ‘1’ had appeared on the die then we say that event E has occurred. As a matter of fact if outcomes are 2 or 3, we say that event E has occurred. Thus, the event E of a sample space S is said to have occurred if the outcome ω of the experiment is such that ω ∈ E. If the outcome ω is such that ω ∉ E, we say that the event E has not occurred.

Types of events:Events can be classified into various types on the basis of the elements they have.

Impossible and Sure Events

The empty set φ and the sample space S describe events. In fact φ is called an impossible event and S, i.e., the whole sample space is called the sure event. To understand these let us consider the experiment of rolling a die. The associated sample space is S = {1, 2, 3, 4, 5, 6}
Let E be the event “ the number appears on the die is a multiple of 7”. Can you write the subset associated with the event E?
Clearly no outcome satisfies the condition given in the event, i.e., no element of the sample space ensures the occurrence of the event E. Thus, we say that the empty set only correspond to the event E. In other words we can say that it is impossible to have a multiple of 7 on the upper face of the die. Thus, the event E = φ is an impossible event.
Now let us take up another event F “the number turns up is odd or even”. Clearly F = {1, 2, 3, 4, 5, 6,} = S, i.e., all outcomes of the experiment ensure the occurrence of the event F. Thus, the event F = S is a sure event.

Simple Event

If an event E has only one sample point of a sample space, it is called a simple (or elementary) event.
In a sample space containing n distinct elements, there are exactly n simple events.
For example in the experiment of tossing two coins, a sample space is S={HH, HT, TH, TT}
There are four simple events corresponding to this sample space. These areE1= {HH}, E2={HT}, E3= { TH} and E4={TT}.

Compound Event

If an event has more than one sample point, it is called a Compound event.
For example, in the experiment of “tossing a coin thrice” the events
G: ‘Atmost one head appeared’ etc.are all compound events.
The subsets of S associated with these events are
E={HTT,THT,TTH}
F={HTT,THT, TTH, HHT, HTH, THH, HHH}
G= {TTT, THT, HTT, TTH}

Each of the above subsets contain more than one sample point, hence they are all compound events.

Algebra of events

Let A, B, C be events associated with an experiment whose sample space is S.

Complementary Event

For every event A, there corresponds another event A′ called the complementary event to A. It is also called the event ‘not A’.

For example, take the experiment ‘of tossing three coins’. An associated sample space is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Let A={HTH, HHT, THH} be the event ‘only one tail appears’

Clearly for the outcome HTT, the event A has not occurred. But we may say that the event ‘not A’ has occurred. Thus, with every outcome which is not in A, we say that ‘not A’ occurs.

Thus the complementary event ‘not A’ to the event A is
A’= {HHH, HTT, THT, TTH, TTT}
or A’= {ω: ω∈S and ω ∉A} = S – A.

The Event ‘A or B’

Recall that union of two sets A and B denoted by A ∪B contains all those elements which are either in A or in B or in both.

When the sets A and B are two events associated with a sample space, then ‘A ∪B’ is the event ‘either A or B or both’. This event ‘A ∪B’ is also called ‘A or B’.
Therefore Event ‘A or B’ = A ∪B = {ω: ω∈A or ω ∈B}

The Event ‘A and B’

We know that intersection of two sets A ∩B is the set of those elements which are common to both A and B. i.e., which belong to both ‘A and B’.

If A and B are two events, then the set A ∩B denotes the event ‘A and B’.
Thus, A ∩B = { ω: ω∈A and ω∈B}

For example, in the experiment of ‘throwing a die twice’ Let A be the event ‘score on the first throw is six’ and B is the event ‘sum of two scores is atleast 11’ then A = {(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}, and B = {(5,6), (6,5), (6,6)} so A ∩B = {(6,5), (6,6)}

Note that the set A ∩B = {(6,5), (6,6)} may represent the event ‘the score on the first throw is six and the sum of the scores is atleast 11’.

The Event ‘A but not B’

We know that A–B is the set of all those elements which are in A but not in B. Therefore, the set A–B may denote the event ‘A but not B’.We know that A – B = A ∩B´

Consider the experiment of rolling a die. Let A be the event ‘getting a prime number’, B be the event ‘getting an odd number’. Write the sets representing the events
(i) Aor B
(ii) A and B
(iii) A but not B
(iv) ‘not A’.

Here S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}
Obviously
(i) ‘A or B’ = A ∪B = {1, 2, 3, 5}
(ii) ‘A and B’ = A ∩B = {3,5}
(iii) ‘A but not B’ = A – B = {2}
(iv) ‘not A’ = A′= {1,4,6}

Mutually exclusive events

In the experiment of rolling a die, a sample space is S = {1, 2, 3, 4, 5, 6}. Consider events, A ‘an odd number appears’ and B ‘an even number appears’

Clearly the event A excludes the event B and vice versa. In other words, there is no outcome which ensures the occurrence of events A and B simultaneously. Here A = {1, 3, 5} and B = {2, 4, 6}
Clearly A ∩B = φ i.e., A and B are disjoint sets.

In general, two events A and B are called mutually exclusive events if the occurrence of any one of them excludes the occurrence of the other event, i.e., if they can not occur simultaneously. In this case the sets A and B are disjoint.

Again in the experiment of rolling a die, consider the events A ‘an odd number appears’ and event B ‘a number less than 4 appears’

Obviously A = {1, 3, 5} and B = {1, 2, 3}
Now 3 ∈ A as well as 3 ∈ B
Therefore, A and B are not mutually exclusive events.

Exhaustive events Consider the experiment of throwing a die. We have S = {1, 2, 3, 4, 5, 6}. Let us define the following events

A: ‘a number less than 4 appears’,
B: ‘a number greater than 2 but less than 5 appears’
and C: ‘a number greater than 4 appears’.
Then A = {1, 2, 3}, B = {3,4} and C = {5, 6}. We observe that A ∪B ∪C = {1, 2, 3} ∪{3, 4} ∪{5, 6} = S.

Such events A, B and C are called exhaustive events

Exhaustive Events:

If n events A1, A2, ........., An related to any particular sample space are such that if we take union of the sets of all the n events, sample space is formed. i.e.
A1 υ A2 υ A3 υ ............... υ An = S
If the union of all event is equal to sample space , the n it is known as exhaustive events

Mutually Exclusive and Exhaustive Events.

Events A1, A2, .......... An are said to be mutually exclusive and exhaustive if they satisfy the condition for mutual exclusion and exhaustiveness both.

i.e. A1 υ A2 υ A3 ............ υ An = S

and Ai ∩ Aj = Φ, where i = 1, 2, ........., n

j = 1, 2, ........... n and i ≠ j.

i.e. mutually exclusive and exhaustive events are those events of which union is equal to sample space and occurrence of any one of them excludes the possibility of occurrence of all others.

Illustration:

In rolling a die, events A1, A2, A3 and A4 are described as follows
A1 : Number less then 2 occurs.
A2 : 2 occurs
A3 : odd number greater that 1 occur.
A4 : even number greater than 2 occur.
A1 : {1}
A2 : {2}
A3 : {3, 5}
A4 : {4, 6}
Clearly, A1 υ A2 υ A3 υ A4 = {1, 2, 3, 4, 5, 6} = S
and A1 ∩ A2 = A1 ∩ A3 = A1 ∩ A4 = A2 ∩ A3 = A2 ∩ A4 = A3 ∩ A4 = Φ
i.e. Ai ∩ Aj = Φ, i ≠ j, i, j = 1, 2, 3, ............ n.

Two dice are thrown and the sum of the numbers which come up on the
dice is noted. Let us consider the following events associated with this experiment
A : ‘the sum is even’.
B : ‘the sum is a multiple of 3’.
C : ‘the sum is less than 4’.
D : ‘the sum is greater than 11’.
Which pairs of these events are mutually exclusive?

There are 36 elements in the sample space S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}.

A coin is tossed three times, consider the following events. A: ‘No head appears’, B: ‘Exactly one head appears’ and C: ‘Atleast two heads appear’.Do they form a set of mutually exclusive and exhaustive events?

The sample space of the experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
and A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH}
Now
A ∪B ∪C = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH} = S
Therefore, A, B and C are exhaustive events.
Also, A ∩B = , A ∩C = and B ∩C =
Therefore, the events are pair-wise disjoint, i.e., they are mutually exclusive.
Hence, A, B and C form a set of mutually exclusive and exhaustive events.

Equally likely events/outcomes

In an experiment, two or more event/outcomes are said to be equally likely, if they have the same chances associated with them. i.e. no one of them has more chance of occurrence than others.
Let S be the sample space & E be the event, such that n(S) = n and n(E) = m.
If each out come is equally likely, then it follows that P(E) = m/n = Number of outcomes favourable to E/ Total possible outcomes

Probability of the event ‘A or B’

Let us now find the probability of event ‘A or B’, i.e., P (A ∪B)
Let A = {HHT, HTH, THH} and B = {HTH, THH, HHH} be two events associated with ‘tossing of a coin thrice’
Clearly A ∪B = {HHT, HTH, THH, HHH}
Now P (A ∪B) = P(HHT) + P(HTH) + P(THH) + P(HHH)
If all the outcomes are equally likely, then
P(A B )= 1/8 + 1/8 +1/8 + 1/8 = 4/8 = ½
Also P(A) = P(HHT) + P(HTH) + P(THH) = 3/8
and P(B ) = P(HTH) + P(THH) + P(HHH) = 3/ 8
Therefore P(A) + P(B ) = 3/8 + 3/8 = 6/8
It is clear that P(A ∪ B ) ≠P(A) + P(B)
The points HTH and THH are common to both A and B . In the computation of P(A) + P(B) the probabilities of points HTH and THH, i.e., the elements of A ∩B are included twice. Thus to get the probability P(A ∪ B ) we have to subtract the probabilities of the sample points in A ∩B from P(A) + P(B )
Thus we observe that, P(A∪B) = P(A) + P(B) − P(A∩B)

Probability of event ‘not A’

Consider the event A = {2, 4, 6, 8} associated with the experiment of drawing a card from a deck of ten cards numbered from 1 to 10. Clearly the sample space is S = {1, 2, 3, ...,10}
If all the outcomes 1, 2, ...,10 are considered to be equally likely, then the probability of each outcome is
1/ 10
Now P(A) = P(2) + P(4) + P(6) + P(8) =1/10 +1/10 +1/10+1/10 = 2/5
Also event ‘not A’ = A′ = {1, 3, 5, 7, 9, 10}
Now P(A′) = P(1) + P(3) + P(5) + P(7) + P(9) + P(10) = 3/5
Thus, P(A′) = 3/5 = 1- 2/5 = 1 P(A)
So P( A′ ) = P(not A) = 1 – P(A)

One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be
(i) a diamond
(ii) not an ace
(iii) a black card (i.e., a club or, a spade)
(iv) not a diamond
(v) not a black card

When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52.

(i) Let A be the event 'the card drawn is a diamond'
Clearly the number of elements in set A is 13.
Therefore, P(A) = 13/52 = 1/4
i.e. Probability of a diamond card = 1/4

(ii) We assume that the event ‘Card drawn is an ace’ is B
Therefore ‘Card drawn is not an ace’ should be B′.
We know that P(B′) = 1 – P(B) = 1-4/52 =12/13

(iii) Let C denote the event ‘card drawn is black card’
Therefore, number of elements in the set C = 26
i.e. P(C) = 26/52 = 1/2
Thus, Probability of a black card = 1/2

(iv) We assumed in (i) above that A is the event ‘card drawn is a diamond’,
so the event ‘card drawn is not a diamond’ may be denoted as A' or ‘not A’
Now P(not A) = 1 – P(A)
1-1/4 = 3/4

(v) The event ‘card drawn is not a black card’ may be denoted as C′or ‘not C’.
We know that P(not C) = 1 – P(C)
1-1/2 =1/2

A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be
(i) red
(ii) yellow
(iii) blue
(iv) not blue
(v) either red or blue.

There are 9 discs in all so the total number of possible outcomes is 9.
Let the events A, B, C be defined as
A: ‘the disc drawn is red’
B: ‘the disc drawn is yellow’
C: ‘the disc drawn is blue’.

(i) The number of red discs = 4, i.e., n (A) = 4
Hence P(A) = 4/9

(ii) The number of yellow discs = 2, i.e., n (B) = 2
Therefore, P(B) = 2/9

(iii) The number of blue discs = 3, i.e., n(C) = 3
Therefore, P(C) = 3/9 =1/3

(iv) Clearly the event ‘not blue’ is ‘not C’. We know that P(not C) = 1 – P(C)
Therefore P(not C) =1- 1/3 = 2/3

(v) The event ‘either red or blue’ may be described by the set ‘A or C’
Since, A and C are mutually exclusive events, we have
P(A or C) = P (A ∪C) = P(A) + P(C) = 4/9 +1/3 = 7/9

Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that
(a) Both Anil and Ashima will not qualify the examination.
(b) Atleast one of them will not qualify the examination and
(c) Only one of them will qualify the examination

Let E and F denote the events that Anil and Ashima will qualify the examination, respectively. Given that
P(E) = 0.05, P(F) = 0.10 and P(E ∩F) = 0.02.

Then

(a) The event ‘both Anil and Ashima will not qualify the examination’ may be expressed as E´ ∩F´.
Since, E´ is ‘not E’, i.e., Anil will not qualify the examination and F´ is ‘not F’, i.e.,
Ashima will not qualify the examination.
Also E´ ∩F´ = (E ∪F)´ (by Demorgan's Law)
Now P(E ∪F) = P(E) + P(F) – P(E ∩F)
or P(E ∪F) = 0.05 + 0.10 – 0.02 = 0.13
Therefore P(E´ ∩F´) = P(E ∪F)´ = 1 – P(E ∪F) = 1 – 0.13 = 0.87

(b) P (atleast one of them will not qualify)
= 1 – P(both of them will qualify)
= 1 – 0.02 = 0.98

(c) The event only one of them will qualify the examination is same as the event
either (Anil will qualify, and Ashima will not qualify) or (Anil will not qualify and Ashima
will qualify) i.e., E ∩F´ or E´ ∩F, where E ∩F´ and E´ ∩F are mutually exclusive.
Therefore, P(only one of them will qualify) = P(E ∩F´ or E´ ∩F)
= P(E ∩F´) + P(E´ ∩F) = P (E) – P(E ∩F) + P(F) – P (E ∩F)
= 0.05 – 0.02 + 0.10 – 0.02 = 0.11

A committee of two persons is selected from two men and two women. What is the probability that the committee will have
(a) no man?
(b) one man?
(c) two men?

The total number of persons = 2 + 2 = 4. Out of these four person, two can be selected in 4 C2 ways.

(a) No men in the committee of two means there will be two women in the committee.
Out of two women, two can be selected in 2 C2 =1 way.
Therefore P( no man) = 2c2/4c2 = 1/6

(b) One man in the committee means that there is one woman. One man out of 2 can be selected in 2
C1 ways and one woman out of 2 can be selected in 2C1 ways.
Together they can be selected in 2C1 × 2C1 ways.
Therefore P (One man ) = 2c1 * 2c1/ 4c2 = 2/3

(c) Two men can be selected in 2C2 way.
Hence P (Two men) 2c2/4c2 = 1/6

On her vacations Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits
(i) A before B?
(ii) A before B and B before C?
(iii) A first and B last?
(iv) A either first or second?
(v) A just before B?

The number of arrangements (orders) in which Veena can visit four cities A, B, C, or D is 4! i.e., 24.Therefore, n (S) = 24.

Since the number of elements in the sample space of the experiment is 24 all of these outcomes are considered to be equally likely. A sample space for the experiment is

(i) Let the event ‘she visits A before B’ be denoted by E
Thus P(E) = n(E)/n(S) = 12/24 =1/2

(ii) Let the event ‘Veena visits A before B and B before C’ be denoted by F.
Here F = {ABCD, DABC, ABDC, ADBC}
Thus P(f) = n(f)/n(s) = 4/24 = 1/ 6

(iii) 2/24 = 1/12

(iv) 12/24 = 1/2

(v) 6/24 = 1/4

I have written all the cases, but you all can solve directly also.

let say 5 person a,b,c,d,e can be arranged in 5! ways ; no. of ways in which a is before b = 5!/2!; no. of ways in which a is before b & b is before c = 5!/3! ; no. of ways in which a is before b & b is before c & c is before d = 5!/4! ; no. of ways in which a is before b & b is before c & c is before d & d is before e = 5!/5! = 1;

Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains
(i) all Kings
(ii) 3 Kings
(iii) atleast 3 Kings.

Total number of possible hands = 52 C7
(i) Number of hands with 4 Kings = 4 C4 × 48C3 (other 3 cards must be chosen from the rest 48 cards)
Hence P (a hand will have 4 Kings) = 4 C4 × 48C3 / 52 C7 = 1/7735

(ii) Number of hands with 3 Kings and 4 non-King cards = 4C3 × 48C4
Therefore P (3 Kings) =4C3 × 48C4 / 52c7 = 9/1547

(iii) P(atleast 3 King) = P(3 Kings or 4 Kings)
= P(3 Kings) + P(4 Kings) = 9/1547 + 1/7735 = 46/7735

In a relay race there are five teams A, B, C, D and E.
(a) What is the probability that A, B and C finish first, second and third respectively.
(b) What is the probability that A, B and C are first three to finish (in any order) (Assume that all finishing orders are equally likely)

If we consider the sample space consisting of all finishing orders in the first three places, we will have 5 P3 = 60 sample points, each with a probability of 1/60.

(a) A, B and C finish first, second and third, respectively. There is only one finishing order for this, i.e., ABC.
Thus P(A, B and C finish first, second and third respectively) = 1/60

(b) A, B and C are the first three finishers. There will be 3! arrangements for A, B and C. Therefore, the sample points corresponding to this event will be 3! In number.
So P (A, B and C are first three to finish) = 3!/60 = 1/10

In a given race the odds in favour of four horses A, B, C, D are 1:3, 1:4, 1:5, 1:6 respectively. Assuming that, a dead heat is impossible, find the chance that one of them wins the race.

Let P(A), P(B), P(C) and P(D) be the responsibilities of winning of the horses A, B, C and D respectively. Then
P(A) = 1/4, P(B) = 1/5, P(C) = 1/6, P(D) = 1/7.
Since the above events are mutually exclusive, the chance that one of them wins
= P(AUBUCUD) = P(A) + P(B) + P(C) + P(D)
= (1/4) + (1/5) + (1/6) + (1/7) .

You can refer Part 2 for the concepts related to Conditional probability & Baye's theorem

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