**on Mutually Exclusive Events**

Q1. If sample space ={1,2,3,…,9}, event

*A*={2,4,6,8} and Event

*B*={1,3,5} find P(

*A∪B*).

Solution:

P(

*A∪B*)=P(

*A*)+P(

*B*) … (i).

*A*)=4/9;P(

*B*)=3/9

Put in equation (i), we get

Q2. Two dice are thrown. What is the probability that the sum of the number of dots appearing on them is 4 or 6?

Solution:

When two dice are rolled then the possible outcomes are

Let

*A*= be the event the sum is 4, then

Let

*B*be the event that the sum is 6, then

Q3. There are 10 girls and 20 boys in a class. Half of the boys and half of the girls have blue eyes. Find the probability that one student chosen as monitor is either a girl or has blue eyes.

Solution:

Q4. A DVD shop has 180 comedies, 250 drama films, 230 science fiction movies and 120 thrillers. If you select a DVD at random, what is the probability that this movie is a comedy OR a thriller?

Solution:

No DVD is marked as both a comedy and a thriller, so there is no overlap in events. These are mutually exclusive (but not complementary).

There are 250+230+120=600 DVDs in the sample space.

Use P(

*A*or

*B*)=P(

*A*)+P(

*B*).

P(comedy or thriller)=P(comedy)+P(thriller)

Q5. 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 in all so the total number of possible outcomes is 9. Let the events

*A*,

*B*,

*C*be defined as

A: ‘the disc drawn is red’

B: ‘the disc drawn is yellow’

C: ‘the disc drawn is blue’.

(i). The number of red discs =4, i.e., n(

*A*)=4

Hence P(

*A*)=4/9

(ii). The number of yellow discs =2, i.e., n(

*B*)=2

Therefore, P(

*B*)=2/9

(iii). The number of blue discs =3, i.e., n(

*C*)=3

Therefore,

P(

*C*)=3/9=⅓

(iv). Clearly the event ‘not blue’ is ‘not

*C*’. We know that

P(not

*C*)=1-⅓=⅔

(v). The event ‘either red or blue’ may be described by the set ‘

*A*or

*C*’. Since,

*A*and

*C*are mutually exclusive events, we have

**on Not Mutually Exclusive Events**

Let’s read the previous posts

Probability of Either Event A or B happens, or Both happen

In the previous posts we demonstrated the

**addition law**of probability:

For two events

*A*and

*B*,

*A∪B*)=P(

*A*)+P(

*B*)-P(

*A∩B*)

which means:

**either**

*A*

**or**

*B*

**or**both) =P(

*A*)+P(

*B*)-P(

**both**

*A*

**and**

*B*)

*X*is 0.43 and the probability of event

*Y*is 0.24. The probability of both occurring together is 0.10. What is the probability that

*X*or

*Y*will occur?

Solution:

From the addition rule

*X*or

*Y*)=P(

*X*)+P(

*Y*)-P(

*X*and

*Y*)

=0.43+024-0.10=0.57

solution:

Step 1: Identify the identity which describes the situation

Step 2: Calculate the answer

Notice how we have used P(

*C∪A*)=P(

*C*)+P(

*A*)-P(

*C∩A*)

Q8. Two dice are rolled; find the probability of getting doubles or a sum of 6.

Solution:

Let *A*= getting doubles; then P(*A*)=6/36 since there are six ways to get doubles and let *B* = getting a sum of 6. Then P(*A*)=5/36 since there are five waysto get a sum of 6→ (5, 1), (4,2), (3,3), (2,4), and (1,5). Let *A*= and *B*=thenumber of ways to get a double and a sum of 6. There is only one way forthis event to occur→namely (3,3); then P(*A* and *B*)=1/36. Hence,

Q9:

From 100 cards, numbered from 1 to 100, one is selected at random. Find the probability that the card selected is even or less than 20.

Solution:

Some cards are both even and less than 20 (i.e. 2, 4, 6, 8, 10, 12, 14, 16, 18).

P(even or <20)=P(even)+P (<20)-P(even and <20)

Q10. A card is drawn from a deck of 52 playing cards what is the probability that it is a diamond card or an ace:

Solution:

Here n(S)=52

Q11. From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows:

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years?

Solution:

Let

*E*be the event in which the spokesperson will be a male and

*F*be the event in which the spokesperson will be over 35 years of age.

Accordingly, P(

*E*)=⅗ and P(

*F*)=⅖

Since there is only one male who is over 35 years of age,

*E∩F*)=⅕

We know that

*E∪F*) =P(

*E*)+P(

*F*)-P(

*E∩F*)

P(

*E∪F*)=⅗+⅖-⅕=⅘

Thus, the probability that the spokesperson will either be a male or over 35 years of age is ⅘.

Let’s red the post Calculating Probability Without Two-Circle Venn Diagram (part 1)

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