Probability is the chance of the outcome of an event of a particular experiment. An event is one or more possible outcomes of an experiment. Probabilities occurs always in numbers between 0(impossible) and 1(possible). The set of all possible outcomes of a particular experiment is called sample space. The complementary event of an event A is denoted as $\bar A$ or A' or A^{c}.

Compliment of complimentary event is again the same event. If A is an event then,

$\bar{\bar A}$ = A.

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Complimentary events are events that can't occur at the same time. The event A and its complement, A^{c} are mutually exclusive and exhaustive. The complementary event A^{c} to an event A is the event 'A does not occur'. It satisfies A ⋃ A^{c }= S, where S is the sample space, and A ∩ A^{c} = ϕ, where ϕ is the empty set.The complement of an event A is the set of all outcomes in the sample space that are not included in the outcomes of event A. The complement of event A can be represented by A^{c} or A' or $\bar A$.

*For example: ***Turning a Head and Turning a Tail are complementary events when a coin is tossed.**

The probability of an event A with regard to a sample space S of an experiment should lie between 0 and 1. The probability of the event S is 1. If in an experiment an event is described as success, then the complementary event can be described as failure. If the probability of success is p, then ‘q’ the probability of failure which is the complement of success is 1 - p. If an experiment is repeated, then we can expect a success or failure in each trial. If the outcome of each trial is independent, we can apply the multiplication rule for independent events to solve some of the probability problems.

The event the tested part is not defective is the complementary event of the part being defective.

Probability of the item being defective P(A) = 0.005

Probability of the item being defective P(A) = 0.005

Probability of the item not being defective P($\bar A$) = 1 - 0.005 = 0.995

[ P(A) + P($\bar A$) = 1 ]

**step 2:**

[ P(A) + P($\bar A$) = 1 ]

If the 5th piece checked is to be the first defective one, then the first four tested parts should be non defective ones.

=> P(Non defective ∩ Non defective ∩ Non defective ∩ Non defective ∩ defective) = (0.995)^{4} x (0.005)

≈ 0.0049

The probability of finding the first non defective in the 5th check = 0.0049.≈ 0.0049

Red balls = 12

Blue balls = 10

=> Total number of balls = 12 + 10 = 22 balls

Two balls can be drawn from bag out of 22 balls = C(22, 2) ways

=> C(22, 2) = $\frac{22!}{2!(22 - 2)!}$

= $\frac{22 * 21 * 20!}{2 * 1 * 20!}$

= 231

Let A be the event that both the balls are red.

The number of ways of selecting 2 red balls out of 12 = C(12, 2)

= $\frac{12!}{2!(12 - 2)!}$

= $\frac{12 * 11 * 10!}{2 * 1 * 10!}$

= 66

Step 3:

Therefore the probability of both the balls are red.

=> P(A) = $\frac{66}{231}$

= $\frac{2}{7}$

Hence the probability of both the balls are red is $\frac{2}{7}$.

Bernoulli trial is an experiment with two outcomes. Bernoulli trials consist of repeated performance of an experiment with only two outcomes. If one of the outcomes is termed as the success, then the other can be called the failure. As these two events are mutually exclusive and exhaustive, they are complementary events. If the probability of success is known as ‘p’ in a single trial, then the probability of failure ‘q’ which is the complementary event of success and is given by 1 - p. The outcome of each of Bernoulli trials is independent. The probabilities of success and failure remain the same for each trial.

The probability of one of the complementary events is success ‘p’ and the number of Bernoulli trials thus form the parameters for the Binomial Probability distribution. The formula for finding the probability of ‘x’ number of successes in trials is given by the formula,

The probability of one of the complementary events is success ‘p’ and the number of Bernoulli trials thus form the parameters for the Binomial Probability distribution. The formula for finding the probability of ‘x’ number of successes in trials is given by the formula,

where, q = 1 - p.

Two events are described as complementary if they are the only two possible outcomes. Two mutually exclusive events that are all inclusive, so they create a sample sample space when combined.

**The sum of the probabilities of complementary events is 1. **

If A is an event, and A' is the complementary event,

P(A) + P($\bar A$) = 1

or

P($\bar A$) = 1 - P(A)

The property P(A) = 1 – P($\bar A$) is effectively used in calculating probabilities in binomial distribution. For many situations which we come across in problems, finding the probability of the complementary event would be a much easier task than finding the required probability.

If A is an event, and A' is the complementary event,

P(A) + P($\bar A$) = 1

or

P($\bar A$) = 1 - P(A)

The property P(A) = 1 – P($\bar A$) is effectively used in calculating probabilities in binomial distribution. For many situations which we come across in problems, finding the probability of the complementary event would be a much easier task than finding the required probability.

The binomial distribution formula for getting exactly x successes in trials is,

P(x) = C(n, x)p^{x}q^{n - x}

where p is the probability of success in a single trial and q = 1 - p.

**Step 2:**

P(x) = C(n, x)p

where p is the probability of success in a single trial and q = 1 - p.

To
find the probability that the student gets at least one question
correct. This means he can get 1 or 2 or 3 or…..........or 8 or 9 or
all 10 questions correct. If we are to use the formula direct, we need
to find the probabilities for x taking all the values 1 to 10 separately
and add all the probabilities we get for exact successes. This would
be a tedious job.

Step 3:

Step 3:

Instead of step 2,
we can make use of complementary event to simplify solving this
problem. The event of getting ‘at least one question correct’ is the
complement of the event ‘getting no question correct’. We can compute
the probability of this event easily, and using the rule P(A) = 1 - P(A), the required probability can be found.

**Step 4:**

Since each question has 4 choices, the probability of success in one trial p = $\frac{1}{4}$ and q = 1 - p = $\frac{3}{4}$. There are 10 questions in the test, hence n =10.

=> Probability (getting no question correct) = P(x = 0)

= C(10,0)($\frac{1}{4})^0$ $(\frac{3}{4})^{10}$ ≈ 0.0563.

= C(10,0)($\frac{1}{4})^0$ $(\frac{3}{4})^{10}$ ≈ 0.0563.

The probability of the complementary event, getting at least one question correct = 1- 0.0563

= 0.9437.** answer**

Given below are some of the examples on Complementary Events:

We know, the sum of probabilities of complimentary events is 1.

Here P(A) = $\frac{5}{9}$

P(A') = 1 - P(A) = 1 - $\frac{5}{9}$ = $\frac{4}{9}$

=> Probability of the complimentary event of the event, P(A') = $\frac{4}{9}$.

Let S be the sample space, n(S) = {1, 2, 3, 4, 5, 6} = 6

A be the event of getting even number.

n(A) = {2, 4, 6} = 3

P(A) = $\frac{(n(A)}{n(B)}$

= $\frac{3}{6}$

= $\frac{1}{2}$

Complimentary event A' = 1 - P(A) = 1 - $\frac{1}{2}$ = $\frac{1}{2}$.

Therefore, the probability of not getting even number is $\frac{1}{2}$.

Let S be the sample space = 6 + 7 = 13

A be the event of getting 2 dark chocolates.

n(S) = C(13,2) = 78

n(A) = C(6,2) = 15

P(A) = $\frac{15}{78}$ = $\frac{5}{26}$

Complimentary event A' = 1 - P(A) = 1 - $\frac{5}{26}$ = $\frac{21}{26}$ .

The event of picking a number not divisible by 10 is the complementary event of picking a number divisible by 10.

100 is the highest number in the slips which is divisible by 10. 10 x 10 = 100. Hence there are 10 numbers from 1 to 100 which are divisible by 10.

Probability of picking a number divisible by 10 = $\frac{10}{100}$ = $\frac{1}{10}$.

The probability of picking a number which is not divisible by 10 = 1 - $\frac{1}{10}$ = $\frac{9}{10}$.

The probability of picking a number which is not divisible by 10 = 1 - $\frac{1}{10}$ = $\frac{9}{10}$.

**Answer: (1) **$\frac{1}{3}$ **(2) **$\frac{33}{95}$

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