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The likelihood of an event to occur or the extent to which the event is probable is called probability. There are many events which cannot be predicted with total certainty, but we can only expect the chance of the event to occur. The likelihood of an event ranges from 0 to 1, where 0 means the event to be an impossible one, and one represents the event to be certain. If we have to calculate the probability of a single event to occur, we should know the total number of possible outcomes. Suppose, the probability of an event E happening is a number P(E), then:

0 ≤ P(E) ≤ 1

The likelihood of all events in a sample space adds up to 1.

Suppose an event E occurs, then the probability of that event to occur P(E) is:

P(E) = The ratio of the number of favourable outcomes by the total number of outcomes.

The basic formula is shown above, but there are more formulas in different probability questions which arise due to different situations and events.

Suppose we toss two coins in the air, then four outcomes can happen, i.e. (H, H), (H, T), (T, H), (T, T). Now, if the probability of an event E containing two heads is asked, then:

P(E) = 1/4, since there is only one case where there are two heads, and there are in total of four outcomes.

There are two types of events:

Impossible Event: The events in which the probability is 0, such events are called impossible events.

Sure Event: When the probability of an event to occur is 1, that is the event certain to happen, such events are called sure events.

These are some probability questions with its solutions:

1. Two coins are tossed 700 times, and the results are:

Two heads: 205 times

One head: 325 times

No head: 170 times

Find the probability of each event to occur.

Solution: Suppose that the events of getting no head, one head and two heads by E1, E2 and E3, respectively.

P(E1) = 170/700 = 0.24

P(E2) = 325/700 = 0.47

P(E3) = 205/700 = 0.29

Since, the sum of probabilities of all events of a random experiment is 1, therefore:

P(E1) + P(E2) + P(E3) = 0.24 + 0.47 + 0.29 = 1

2. If P(A) = 7/13, P(B) = 9/13 and P(A∩B) = 4/13, evaluate P(A|B).

Solution:

Here, P( A ), P( B ) and P(A∩B) is given, we know that –

P(A|B) = (A∩B)/ P(B). Therefore:

(4/13) / (9/13) = 4/9

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These are some solved examples of probability questions as follows:

1. Two dice are rolled, find the probability that the sum of the number of both the dice is:

Equal to 2

Equal to 5

Less than 12

Solution:

S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Total number of possible outcomes, i.e. the sample space = 36

1) Let A be the event of the sum being equal to 2. Since, there is only one outcome where the sum is equal to 2, hence,

P(A) = ratio of n(A) and n(S) = n(A)/n(S) = 1 / 36

2) Let B be the event of the sum being equal to 5. There are four possible outcome that give a sum equal to 5, they are:

B = {(1,4), (2,3), (3,2), (4,1)}

Hence, P(B) = n(B) / n(S) = 4 / 36 = 1 / 9

3) Let C be the event of the sum being less than 12. From the sample space, we can see all possible outcomes for event C, which give a sum less than 12. Like:

(1,1) or (1,6) or (2,6) or (5,6).

We can see that there are 35 outcomes in which the sum is less than 12. Hence,

P(C) = n(C) / n(S) = 35/36.

2. Calculating the probability of selecting a black card or a six from a deck of 52 cards.

Solution:

Let B be the event of selecting a black card.

We need to find out P(B or 6)

Selecting a black card P(B) = 26/52 = 1/2

Selecting a 6 P(6) = 4/52

The probability of selecting both a black card and a 6 = 2/52

We know that-

P(B ∪ 6) = P(B) + P(6) – P(B ∩ 6)

Therefore

= 26/52 + 4/52 – 2/52

= 28/52

= 7/13.

FAQ (Frequently Asked Questions)

1. What are the Different Types of Probability?

Ans: There are three major types of probabilities:

**Theoretical Probability:** In this type of possibility, we assume that all n possible outcomes of a particular experiment are likely to equal, and we assign a probability of 1/n to each of the possible results.

**Experimental Probability:** This type of possibility is determined based on the results of an experiment repeated many times. Empirical probability is based on observation or actual measurements.

**Subjective Probability:** Subjective possibility is derived from an individual's past experience or opinion about whether a specific result of an event is likely to occur or not. It only reflects an individual's judgement.

2. What are the Basic Rules of Probability?

Ans: The basic rules of probability are:

The addition rule will apply when there is a union of 2 other events. Suppose A and B are two events, then:

P(A∪B ) = P(A) + P(B) − P(A∩B)

The complementary rule will apply whenever an event is a complement of another event. Let A be the event whose complement is to be found:

P(A̅) = 1 – P(A)

The conditional probability is applied whenever partial knowledge about an event is available. Let A be the event known, and the probability of B is desired, then:

P(B/A) = P(A∩B)/P(A)

The multiplication rule is applied whenever an event is an intersection of two other events. Suppose A and B occur simultaneously, then:

P(A∩B) = P(B) ⋅ P(A|B)