From Example 1, we note that
`P(E) + P(F) = 1/2+1/2 = 1` ..........(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) = 1/3+2/3 = 1` ............(2)
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 (1) and (2) above, is F not the same as ‘not E’? Yes, it is. We denote the event ‘not E’ by `barE` .
So, `P(E) + P(not E) = 1`
i.e., `P(E) + P( E ) = 1`, which gives us` P( E ) = 1 – P(E).`
In general, it is true that for an event `E,`
`P(barE ) = 1 – P(E)`
The event `barE` , representing ‘not `E`’, is called the complement of the event `E`. We also say that `E` and `barE` 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 favourable 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.
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 favourable 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 favourable to the event E) is always less than or equal to the denominator (the number of all possible outcomes). Therefore,
`0 ≤ P(E) ≤ 1`
Now, let us take an example related to playing cards. Have you seen a deck of
playing cards? It consists of 52 cards which are divided into 4 suits of 13 cards each—
spades (♠), hearts (), diamonds (⧫) and clubs (¨♣). Clubs and spades are of black
colour, while hearts and diamonds are of red colour. The cards in each suit are ace,
king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings, queens and jacks are called face
cards.
From Example 1, we note that
`P(E) + P(F) = 1/2+1/2 = 1` ..........(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) = 1/3+2/3 = 1` ............(2)
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 (1) and (2) above, is F not the same as ‘not E’? Yes, it is. We denote the event ‘not E’ by `barE` .
So, `P(E) + P(not E) = 1`
i.e., `P(E) + P( E ) = 1`, which gives us` P( E ) = 1 – P(E).`
In general, it is true that for an event `E,`
`P(barE ) = 1 – P(E)`
The event `barE` , representing ‘not `E`’, is called the complement of the event `E`. We also say that `E` and `barE` 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 favourable 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.
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 favourable 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 favourable to the event E) is always less than or equal to the denominator (the number of all possible outcomes). Therefore,
`0 ≤ P(E) ≤ 1`
Now, let us take an example related to playing cards. Have you seen a deck of
playing cards? It consists of 52 cards which are divided into 4 suits of 13 cards each—
spades (♠), hearts (), diamonds (⧫) and clubs (¨♣). Clubs and spades are of black
colour, while hearts and diamonds are of red colour. The cards in each suit are ace,
king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings, queens and jacks are called face
cards.