# Probability

In our daily life we come across many situations in which we have to tell the certainty of certain events but we can’t tell 100% sure that event may occur or not. In such situations we have tell the probability of that event that means it is not the exact value but approximate value of occurring the event.

So many things in our life happening not definitely that means they are happening with some probability. In short, probability is the guessing of the event it occurs or not when we are not 100% sure about it.

In mathematics, most of the time we use probability when one or more events are occurring simultaneously and we have to find the probability of occurring one particular event.

Thus, probability has great importance not only in mathematics but also in our daily life. In this article we see in detail how probability can be calculated.

## Probability

• If some event is occurring then probability of occurring that event in mathematical form is given by,

P (event) = no. of successful outcomes/ no. of possible outcomes

• If we shall note that, in which range the value of probability of some event takes value, then we say that it doesn’t happens or happens to a certain value or happens definitely.
• Thus, we can say that, the probability of any event takes value between 0 to 1 only.

Hence,  0 ≤ P (Event) ≤ 1

• And the sum of probabilities of all events in a sample space is always 1.

That means, P(E1) + P(E2) + P(E3) +….= 1

Note: sample space is the set which include all the events

• If A is an event and Ac is the compliment of event A then the probability of compliment of event Ac is given by,

P(Ac) = 1 – P(A)

### Fundamental Principal of Counting:

If one event is occurring in ‘m’ ways and another event is occurring in ‘n’ ways then these both events may occur in (m*n) ways.

### Tossing of coins:

• When one coin is tossed:

Sample space S = {H, T} and n(S) = 2

• When two coins are tossed:

Sample space S = {HH, HT, TH, TT} and n(S) = 4

• When three coins are tossed:

Sample space S = {HHH, HHT, HTH, THH, HTT, TTH, THT, TTT} and n(S)= 8

Thus, in short by fundamental principal of counting,

• 1 coin tossed: n(S) = 2
• 2 coins tossed: n(S) = 2*2 = 4
• 3 coins tossed: n(S) = 2*2*2 = 8
• 4 coins tossed: n(S) = 2*2*2*2 = 16 and so on

Note:

• At least one: one and more than one be acceptable i.e.1 &> 1
• At most one: zero and only less than one be acceptable i.e.1 &< 1

### Solved examples:

1.) When two coins are tossed then find the probability of finding at least one head.

Ans:

When two coins are tossed:

Sample space S = {HH, HT, TH, TT} and n(S) = 4

Now, at least one head means minimum one head and maximum any are allowed.

Event A = {HH, HT, TH} and n(A) = 3

Thus, P(A) = ¾

2.) When two coins are tossed then find the probability of finding at most one head.

Ans:

When two coins are tossed:

Sample space S = {HH, HT, TH, TT} and n(S) = 4

Event A = {HT, TH, TT} and n(A) = 3

Thus, P(A) = ¾

3.) When three coins are tossed find the probability of finding at least three heads.

Ans:

When three coins are tossed:

Sample space S = {HHH, HHT, HTH, THH, HTT, TTH, THT, TTT} and n(S)= 8

Now, at least three heads and maximum any are allowed.

Event A = {HHH} and n(A) = 1

Thus, P(A) = 1/8

### Tossing of  dice:

• When one dice is thrown:

Sample space S = {1, 2, 3, 4, 5, 6} and n(S) = 6

• When two dice are thrown:

Sample space 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)}

And n(S) = 36

Thus, in short

• 1 dice thrown: n(S) = 6
• 2 dice thrown: n(S) = 6*6 = 36
• 3 dice thrown n(S) = 6*6*6 = 216

### Solved examples:

1.) When two dice are thrown then find the probability of getting same numbers.

Ans:

When two dice are thrown:

Sample space 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)}

And n(S) = 36

Now, to find probability of getting same numbers on dice.

Event A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} and n(A) = 6

Thus, P(A) = 6/36 = 1/6

2.) When two dice are thrown then find the probability of getting the product becomes perfect square.

Ans:

When two dice are thrown:

Sample space 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)}

And n(S) = 36

Now, we have to find the probability of getting the product of two number must be perfect square which may be 1, 4, 9, 16, 25, 36 and so on.

Event A = {(1,4), (4,1), (2,2), (1,1), (3,3), (4,4), (5,5), (6,6)}

Hence, n(A) = 8

Thus, P(A) = 8/36 = 2/9

3.) When two dice are thrown then find the probability of getting the product equal to 12.

Ans:

When two dice are thrown:

Sample space 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)}

And n(S) = 36

Now, we have to find the probability of getting the product of numbers equal to 12.

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

Thus, P(A) = 4/36 = 1/9

Note:

The total probability is always one.

Updated: September 9, 2021 — 12:32 pm