Complimentary Events

The event ‘E’ and the
event ‘not E’ are called complementary event of the latter
event. If E occurs, its compliment is E which does not
occur.

Compliment of an event is denoted by E’ or E or E^c.

For example:

1. When a coin is tossed, getting ‘head’ and getting ‘tail’ are complimentary event of each other.

2. When two coins are tossed, getting ‘atleast one head’ and getting ‘no head’ are complimentary event of each other.

3. When a die is thrown:

● Getting ‘even face’ and ‘odd face’ are complimentary event of each other.

● Getting ‘multiple of 2’ and getting ‘not multiple of 2’ are complement event of the latter event.

● Getting ‘divisible by 3’ and getting ‘not divisible by 3’ are complement event of the latter event.

Sum of the probability of an event ‘E’ and probability of an event ‘not E’ is always 1.

i.e., P(E) + P(not E) = 1

Therefore, P = 1 – P(not E)

Or, P(not E) = 1 – P(E)

 

Now we will solve the examples on different types of word problems on complementary event.

Problems on complementary event:

1. A bag contains red and what
balls. The probability of getting a red
ball from the bag of balls is 1/6. What is the probability of not
getting a red ball?

Solution:

The
probability of getting a red ball from the bag of balls is 1/6.

Therefore,
the probability of not getting a red ball

P(ball
is not red) = 1 – 1/6 = 5/6

Therefore, the probability of not getting a
red ball is 5/6.

 

2. In
a box, contains blue and green marbles. The probability of getting a green marble from the box of marbles is 3/7.
What is the probability of getting a blue marble?

Solution:

Let E₁ be the event of getting a green marble and

E₂ be the event of getting a blue marble

E₂ is the probability of getting a blue marble which is also the same as the probability of not getting a green marble, Since we know that the marble are either green or blue.

Therefore, P(not getting a blue marble)

= P(E₂) = 1 – P(E₁)

= 1 – 3/7

= 4/7.

Therefore, the probability of getting a blue marble is 4/7.

3. In a cricket
tournament Yuvraj Singh hits eight times ‘6’ out of thirty two balls. Calculate
the probability that he would not hit a 6?

Solution:

Let P(A) = 32be the
total number of events .

Favorable events that is Yuvraj Singh hits a boundary
P (B) = 8,

Therefore, P(E)
= P (Yuvraj Singh hits a ‘6’)

= P(B)/P(A)

= 8/32 = ¼.

Now, P (not E) = P (Yuvraj Singh did not hit a ‘6’) =
1 – 1/4 = 3/4.

 

4.
In a laptop shop there are 16 defective laptops out of 200 laptops. If one
laptop is taken out at random from this laptop shop, what is the probability
that it is a non defective laptop?

Solution:

The
total number of laptops in laptop shop = 200,

The number of defective laptops = 16,

Let E₁be the event of getting a defective laptops and
E₂ be the event of getting a non defective laptops

P(A) = The probability of getting a defective laptop

= 16/200

= 0.08

Therefore, the probability of getting a non defective
laptop = 1 – P(A) = 1 – 0.08 = 0.92.

 

5.
The probability that it will rain in the evening 0.84. What is the probability
that it will not rain in the evening?

Solution:

Let E be the event that it will rain in the evening.

Then, (not E) is the event it will rain in the evening
.

Then, P(E) = 0.84

Now, P(E) + P(not E) = 1

⇒ P(not E) = 1 – P(E)

⇒ P(not E) = 1 – 0.84

⇒ P(not E) = 0.16

Therefore, the probability that it will not rain in
the evening = P(not E) = 0.16