16.7 - Probability of The Event "not A"

In the previous section, we saw how to calculate the Probability of the event "A or B". In this section, we will see probability of the event "not A".

Probability of the event “not A”

This can be explained using an example. It can be written in 5 steps:
1. Consider the following experiment:
A deck of 10 cards is placed on the table. Each card is given a unique number from 1 to 10. One card is drawn at random without looking. If the number on the drawn card is “an even number less than 10”, then event A is said to occur.
2. It is clear that:
   ♦ set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
   ♦ set A = {2, 4, 6, 8}
   ♦ set ‘not A’ = A’ = {1, 3, 5, 7, 9, 10}
3. There are 10 possible outcomes.
• If all outcomes are equally likely, the probability of each outcome in S will be 1/10.
4. Now let us calculate the various probabilities:
• Since all outcomes in S are equally likely, we have:
   ♦ $\rm{P(A) = \frac{n(A)}{n(S)} = \frac{4}{10} = \frac{2}{5}}$
   ♦ $\rm{P(A') = \frac{n(A')}{n(S)} = \frac{6}{10} = \frac{3}{5}}$
5. Now we can write an interesting point:
$\rm{1 - P(A)~=~1 - \frac{2}{5}~=~\frac{3}{5}~=~P(A')}$
• That means:
$\rm{1 - P(A)~=~P(A')}$
If we have the probability P(A) of any event A, then we can find P(A') by subtracting it from 1.

---

We derived the above formula by using an example. We must write the general proof. It can be written in 2 steps:
1. Recall that, union of any set A and it’s compliment will give the universal set.
• In our present case, we can write: A∪A’ = S
2. So we get:
$\begin{array}{ll}
{}&{\rm{A \cup A'}}
& {~=~}& {\rm{S}}
&{} \\

{\Rightarrow}&{\rm{P(A \cup A')}}
& {~=~}& {\rm{P(S)}}
&{} \\

{\Rightarrow}&{\rm{P(A) + P(A') - P(A \cap A')}}
& {~=~}& {\rm{P(S)~\color{green}{\text{- - - I}}}}
&{} \\

{\Rightarrow}&{\rm{P(A) + P(A') - P(\phi)}}
& {~=~}& {\rm{P(S)~\color{green}{\text{- - - II}}}}
&{} \\

{\Rightarrow}&{\rm{P(A) + P(A') - 0}}
& {~=~}& {\rm{P(S)~\color{green}{\text{- - - III}}}}
&{} \\

{\Rightarrow}&{\rm{P(A) + P(A')}}
& {~=~}& {\rm{1~\color{green}{\text{- - - IV}}}}
&{} \\

{\Rightarrow}&{\rm{P(A')}}
& {~=~}& {\rm{1 - P(A)}}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as I:
In this line, we use the formula: P(A∪B) = P(A) + P(B) – P(A∩B)
• Line marked as II:
In this line we apply the fact:
Intersection of any set and it’s compliment will give a null set.
• Line marked as III:
In this line we apply the fact:
Probability of an event with a null set is zero.
• Line marked as IV:
In this line we apply axiom 2: P(S) = 1

---

Now we will see two solved examples:
Solved example 16.10
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.
Solution:
• There are 52 cards. So there are 52 possible outcomes.
• That means set S will have 52 elements. In other words, n(S) = 52.
• Also, all 52 outcomes are equally likely.

Part (i):
1. Let A be the event: getting a diamond.
Then set A will have 13 elements. In other words, n(A) = 13
2. Given that, all outcomes are equally likely. So we get:
$\rm{P(A) = \frac{n(A)}{n(S)} = \frac{13}{52} = \frac{1}{4}}$

Alternate method:
1. All 52 outcomes have the same probability which is $\frac{1}{52}$
2. So $\rm{P(A) = 13 \times \frac{1}{52} = \frac{13}{52} = \frac{1}{4}}$ 

Part (ii):
1. Let B be the event: getting an ace.
Then set B will have 4 elements. In other words, n(B) = 4
2. Given that, all outcomes are equally likely. So we get:
$\rm{P(B) = \frac{n(B)}{n(S)} = \frac{4}{52} = \frac{1}{13}}$
3. The event “not ace” can be denoted as “not B”. So we get:
$\rm{P(not~B) = P(B’) = 1 – P(B) = 1 - \frac{1}{13} = \frac{12}{13}}$

Part (iii):
1. Let C be the event: getting a black card.
Then set C will have 26 elements. In other words, n(C) = 26
2. Given that, all outcomes are equally likely. So we get:
$\rm{P(C) = \frac{n(C)}{n(S)} = \frac{26}{52} = \frac{1}{2}}$

Part (iv):
1. Let D be the event: getting a diamond.
Then set D will have 13 elements. In other words, n(D) = 13
2. Given that, all outcomes are equally likely. So we get:
$\rm{P(D) = \frac{n(D)}{n(S)} = \frac{13}{52} = \frac{1}{4}}$
3. The event “not a diamond” can be denoted as “not D”. So we get:
$\rm{P(not D) = P(D’) = 1 – P(D) = 1 - \frac{1}{4} = \frac{3}{4}}$

Part (v):
1. Let E be the event: getting a black card.
Then set E will have 26 elements. In other words, n(E) = 26
2. Given that, all outcomes are equally likely. So we get:
$\rm{P(E) = \frac{n(E)}{n(S)} = \frac{26}{52} = \frac{1}{2}}$
3. The event “not getting a black card” can be denoted as “not E”. So we get:
$\rm{P(not~E) = P(E’) = 1 – P(E) = 1 - \frac{1}{2} = \frac{1}{2}}$

Solved example 16.11
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.
Solution:
• There are 9 discs. So there are 9 possible outcomes.
• That means set S will have 9 elements. In other words, n(S) = 9.
• Also, all 9 outcomes are equally likely.

Part (i):
1. Let R be the event: getting a red disc.
Then set R will have 4 elements. In other words, n(R) = 4
2. Given that, all outcomes are equally likely. So we get:
$\rm{P(R) = \frac{n(R)}{n(S)} = \frac{4}{9}}$

Part (ii):
1. Let Y be the event: getting a yellow disc.
Then set Y will have 2 elements. In other words, n(Y) = 2
2. Given that, all outcomes are equally likely. So we get:
$\rm{P(Y) = \frac{n(Y)}{n(S)} = \frac{2}{9}}$

Part (iii):
1. Let B be the event: getting a blue disc.
Then set B will have 3 elements. In other words, n(B) = 3
2. Given that, all outcomes are equally likely. So we get:
$\rm{P(B) = \frac{n(B)}{n(S)} = \frac{3}{9} = \frac{1}{3}}$

Part (iv):
• The event “not blue” can be denoted as “not B”. So we get:
$\rm{P(not~B) = P(B’) = 1 – P(B) = 1 - \frac{1}{3} = \frac{2}{3}}$

Part (v):
1. The event "either red or blue" is (R∪B)
2. R and B are mutually exclusive events. So we can write:
$\begin{array}{ll}
{}&{\rm{P(R \cup B)}}
& {~=~}& {\rm{P(R) + P(B)}} &{} \\

{}&{}
& {~=~}& {\rm{\frac{4}{9} + \frac{3}{9}}} &{} \\

{}&{}
& {~=~}& {\rm{\frac{7}{9}}} &{} \\

\end{array}$

---

In the next section, we will see a few more solved examples.

Previous

Contents

Next

Copyright©2023 Higher secondary mathematics.blogspot.com