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).

P(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

n(S)=36
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 – Mutually Exclusive Events or Not
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,

P(A∪B)=P(A)+P(B)-P(A∩B)
which means:

P(either A or B or both) =P(A)+P(B)-P(both A and B)Q6. The probability of event 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

P(X or Y)=P(X)+P(Y)-P(X and Y)
=0.43+024-0.10=0.57Q7. What is the probability of drawing a club or an ace with one single pick from a pack of 52 cards
solution:
Step 1: Identify the identity which describes the situation

P(club ∪ ace)=P(club)+P(ace)-P(club ∩ ace)
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,

P(E∩F)=⅕
We know that

P(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)