Calculating Probability of the Complement of each Event with Combination Formula

Let us find the probability for some of the events associated with experiments where the equally likely assumption holds.

Example 1: Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.
Solution: In the experiment of tossing a coin once, the number of possible outcomes is two — Head (H) and Tail (T). Let E be the event ‘getting a head’. The number of outcomes favorable to E, (i.e., of getting a head) is 1. Therefore,

Similarly, if F is the event ‘getting a tail’, then
P(F)=P(tail)=½(Why?)

Example 2: A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the
(i) yellow ball? (ii) red ball? (iii) blue ball?
solution: Kritika takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them.
Let Y be the event ‘the ball taken out is yellow’, B be the event ‘the ball taken out is blue’, and R be the event ‘the ball taken out is red’.
Now, the number of possible outcomes =3.
(i) The number of outcomes favorable to the event Y=1.
So, P(Y)=⅓
Similarly, (ii) P(R)=⅓ and (iii) P(B)=⅓
Remarks:
1. An event having only one outcome of the experiment is called an elementary event. In Example 1, both the events E and F are elementary events. Similarly, in Example 2, all the three events, Y, B and R are elementary events.
2. In Example 1, we note that: P(E)+P(F)=1
In Example 2, we note that: P(Y)+P(R)+P(B)=1
Observe that the sum of the probabilities of all the elementary events of an experiment is 1. This is true in general also.

Example 3: Suppose we throw a die once. (i) What is the probability of getting a number greater than 4? (ii) What is the probability of getting a number less than or equal to 4?
solution: (i) Here, let E be the event ‘getting a number greater than 4’. The number of possible outcomes is six: 1, 2, 3, 4, 5 and 6, and the outcomes favorable to E are 5 and 6. Therefore, the number of outcomes favorable to E is 2. So,

P(E)=P(number greater than 4)=2/6=⅓

(ii) Let F be the event ‘getting a number less than or equal to 4’.
Number of possible outcomes =6
Outcomes favorable to the event F are 1, 2, 3, 4.
So, the number of outcomes favorable to F is 4.
Therefore, P(F)=4/6=⅔

Are the events E and F in the example above elementary events? No, they are not because the event E has 2 outcomes and the event F has 4 outcomes.
Remarks: From Example 1, we note that

P(E)+P(F)=½+½=1

where E is the event ‘getting a head’ and F is the event ‘getting a tail’.
From (i) and (ii) of Example 3, we also get
P(E)+P(F)=⅓+⅔=1

where E is the event ‘getting a number >4’ and F is the event ‘getting a number 4’.
Note that getting a number not greater than 4 is same as getting a number less than or equal to 4, and vice versa.
In the examples (1) and (3) above, is F not the same as ‘not E’? Yes, it is. We denote the event ‘not E’ by Ē.
So, P(E)+P(not E)=1
i.e., P(E)+P(Ē)=1, which gives us P(Ē)=1-P(E).
In general, it is true that for an event E,
P(Ē)=1-P(E)

The event Ē, representing ‘not E’, is called the complement of the event E.
We also say that E and Ē are complementary events.
Before proceeding further, let us try to find the answers to the following questions:
(i) What is the probability of getting a number 8 in a single throw of a die?
(ii) What is the probability of getting a number less than 7 in a single throw of a die?
Let us answer (i) :
We know that there are only six possible outcomes in a single throw of a die. These outcomes are 1, 2, 3, 4, 5 and 6. Since no face of the die is marked 8, so there is no outcome favorable to 8, i.e., the number of such outcomes is zero. In other words, getting 8 in a single throw of a die, is impossible.
So, P(getting 8)=0/6=0

That is, the probability of an event which is impossible to occur is 0. Such an event is called an impossible event.
Let us answer (ii):
Since every face of a die is marked with a number less than 7, it is sure that we will always get a number less than 7 when it is thrown once. So, the number of favorable outcomes is the same as the number of all possible outcomes, which is 6.
Therefore, P(E)=P(getting a number less than 7)=6/6=1
So, the probability of an event which is sure (or certain) to occur is 1. Such an event is called a sure event or a certain event.
Note: From the definition of the probability P(E), we see that the numerator (number of outcomes favorable to the event E) is always less than or equal to the denominator (the number of all possible outcomes). Therefore,

0≤P(E)≤1

Note: You can also find P(E) as follows:
P(E)=1-P(Ē) (Since P(Ē)=P(no head)=¼)
Did you observe that in all the examples discussed so far, the number of possible outcomes in each experiment was finite? If not, check it now.
There are many experiments in which the outcome is any number between two given numbers, or in which the outcome is every point within a circle or rectangle, etc.
Can you now count the number of all possible outcomes? As you know, this is not possible since there are infinitely many numbers between two given numbers, or there are infinitely many points within a circle. So, the definition of (theoretical) probability which you have learnt so far cannot be applied in the present form. What is the way out? To answer this, let us consider the following examples:

Q1. If we select three numbers out of first 100 natural numbers. then what is the probability that product of these three numbers is even?

correct: a, solution:
The product will be odd if all three are odd whose probability

The probability that the product will be even

Q2. In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets.
solution:
Total number of tickets sold =10,000. Number prizes awarded =10.
(i). If we buy one ticket, then

(ii). If we buy two tickets, then
Number of tickets not awarded =10,000-10=9990

(iii) If we buy 10 tickets, then

Q3. Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that
(a) you both enter the same section?
(b) you both enter the different sections?
solution:
My friend and I are among the 100 students.
Total number of ways of selecting 2 students out of 100 students =100C2
(a). The two of us will enter the same section if both of us are among 40 students or among 60 students.
∴Number of ways in which both of us enter the same section =40C2+60C2.
∴Probability that both of us enter the same section

(b). P(we enter different sections)
=1-P(we enter the same section)

Q4. 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?
solution:
The total number of persons =2+2=4. Out of these four person, two can
be selected in 4C2 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 2C2=1 way or no man will be selected in 2C0=1 way. Therefore

(b). One man in the committee means that there is one woman. One man out of 2 can be selected in 2C1 ways and one woman out of 2 can be selected in 2C1 ways. Together they can be selected in 2C1×2C1 ways. Therefore

(c). Two men can be selected in 2C2 ways. Hence