Probability Definition and Important Examples

Read About: Sample Space and Event in Probability

Also, 0 ≤ P(A) ≤ 1
⇒ P(Ā) + P(A) = 1
⇒ P(Ā) + = 1 – P(A)

Example 1: A die is thrown. Find the probability of getting- (i) a Prime Number (ii) 2 or 4 (iii) a Multiple of 2 or 3 (iv) a number greater than 2 (v) an even prime.

Solution: Let ‘S’ be the sample space of throwing a die.
S = { 1, 2, 3, 4, 5, 6 }
⇒ n(S) = 6

(i) Let A be the event of getting a prime number.
A = { 2, 3, 5 }
⇒ n(A) = 3

∴ P(A) = n(A) / n(S)
⇒ P(A) = 3/6 = 1/2

(ii) Let B be the event of getting 2 or 4.
B = { 2, 4 }
⇒ n(B) = 2

∴ P(B) = n(B) / n(S)
⇒ P(B) = 2/6 = 1/3

(iii) Let C be the event of getting a multiple of 2 or 3.
C = { 2, 4, 6, 3 }
⇒ n(C) = 4

∴ P(C) = n(C) / n(S)
⇒ P(C) = 4/6 = 2/3

(iv) Let D be the event of getting a number greater than 2.
D = { 3, 4, 5, 6 }
⇒ n(D) = 4

∴ P(D) = n(D) / n(S)
⇒ P(D) = 4/6 = 2/3

(v) Let E be the event of getting an even prime.
E = { 2 }
⇒ n(E) = 1

∴ P(E) = n(E) / n(S)
⇒ P(B) = 1/6

Example 2: In a single throw of two dies find the probability of getting- (i) a total of 9 (ii) a total of 11 (iii) 6 as a product (iv) a total of 8 or 10 (v) a doublet (vi) a sum greater than 9 (vii) neither 9 nor 11 as the sum of the numbers on the faces (viii) a number greater than four on each die (ix) neither a doublet nor a total of six will appear.

Solution- Let S be the sample space of throwing two dies.
sample space probability

n(S) = 36

(i) Let A be the event of getting a total of 9.
A = { (3, 6) (4, 5) (5, 4) (6, 3) }
⇒ n(A) = 4

∴ P(A) = n(A) / n(S)
⇒ P(A) = 4/36 = 1/9

(ii) Let B be the event of getting a total of 11.
B = { (5, 6) (6, 5) }
⇒ n(B) = 2

∴ P(B) = n(B) / n(S)
⇒ P(B) = 2/36 = 1/18

(iii) Let C be the event of getting 6 as a product.
C = { (1, 6) (2, 3) (3, 2) (6, 1) }
⇒ n(C) = 4

∴ P(C) = n(C) / n(S)
⇒ P(C) = 4/36 = 1/9

(iv) Let D be the event of getting a total of 8 or 10.
D = { (2, 6) (3, 5) (4, 4) (5, 3) (6, 2) (4, 6) (5, 5) (6, 4) }
⇒ n(D) = 8

∴ P(D) = n(D) / n(S)
⇒ P(D) = 8/36 = 2/9

(v) Let E be the event of getting a doublet.
E = { (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6) }
⇒ n(E) = 6

∴ P(E) = n(E) / n(S)
⇒ P(B) = 6/36 = 1/6

(vi) Let A be the event of getting a sum greater than 9 i.e. 10, 11, and 12.
A = { (4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6) }
⇒ n(A) = 6

∴ P(A) = n(A) / n(S)
⇒ P(A) = 6/36 = 1/6

(vii) Let A be the event of getting a total of either 9 or 11.
A = { (3, 6) (4, 5) (5, 4) (6, 3) (5, 6) (6, 5) }
⇒ n(A) = 6

∴ P(A) = n(A) / n(S)
⇒ P(A) = 6/36 = 1/6

∴ P(neither 9 nor 11 as sum) = 1 – P(either 9 or 11 as sum)
⇒ P(neither 9 nor 11 as sum) = 1 – P(A)
⇒ P(neither 9 nor 11 as sum) = 1 – 1/6 = 5/6

(viii) Let A be the event of getting a number greater than 4 on each die.
A = { (5, 5) (5, 6) (6, 5) (6, 6) }
⇒ n(A) = 4

∴ P(A) = n(A) / n(S)
⇒ P(A) = 4/36 = 1/9

(ix) Let A be the event of getting either a doublet or a total of six.
A = { (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6) (1, 5) (2, 4) (4, 2) (5, 1) }
⇒ n(A) = 10

∴ P(A) = n(A) / n(S)
⇒ P(A) = 10/36 = 5/18

∴ P(neither a doublet nor a total of 6) = 1 – P(either a doublet or a total of 6)
⇒ P(neither a doublet nor a total of 6) = 1 – P(A)
⇒ P(neither a doublet nor a total of 6) = 1 – 5/18 = 13/18

Example 3: A coin is tossed three times. Find the probability of- (i) no head (ii) exactly one head (iii) at least one head (iv) at most one head (v) at least two heads.

Solution- Let S be the sample space of tossing of a coin three times.
∴ S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT }
⇒ n(S) = 8

(i) Let A be the event of getting no head.
A = { TTT }
⇒ n(A) = 1

∴ P(A) = n(A) / n(S)
⇒ P(A) = 1/8

(ii) Let A be the event of getting exactly one head.
A = { HTT, THT, TTH }
⇒ n(A) = 3

∴ P(A) = n(A) / n(S)
⇒ P(A) = 3/8

(iii) Let A be the event of getting at least one head.
A = { HTT, THT, TTH, HHT, HTH, THH, HHH }
⇒ n(A) = 7

∴ P(A) = n(A) / n(S)
⇒ P(A) = 7/8

OR

P(at least one head) = 1 – P(no head)
⇒ P(at least one head) = 1 – 1/8 = 7/8

(iv) Let A be the event of getting at most one head.
A = { HTT, THT, TTH, TTT }
⇒ n(A) = 4

∴ P(A) = n(A) / n(S)
⇒ P(A) = 4/8 = 1/2

(v) Let A be the event of getting at least two heads.
A = { HHT, HTH, THH, HHH }
⇒ n(A) = 4

∴ P(A) = n(A) / n(S)
⇒ P(A) = 4/8 = 1/2

Example 4: What is the probability that a non-leap year selected at random has 53 Sundays?

Solution- In a non-leap year there are 365 days in which there are 52 weeks and 1 day. In 52 weeks there are 52 Sundays. Now 1 day can be any of the seven days. Let S be the sample space associated with the one day i.e.-

S = { Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday }

Out of 7 outcomes, only 1 is favorable i.e. Sunday.

∴ P(53 Sundays) = 1/7

Example 5: What is the probability that a leap year selected at random has 53 Sundays?

Solution- In a leap year there are 366 days in which there are 52 weeks and 2 days. In 52 weeks there are 52 Sundays. Now the remaining 2 days can be any of the following days. Let S be the sample space associated with these two remaining days i.e.-

S = { (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday) }

Let A be the event of 53 Sundays.

A = { (Sunday, Monday), (Saturday, Sunday) }
n(A) = 2
∴ P(A) = 2/7

Example 6: Three dice are thrown together. Find the probability of getting a total of at least 6.

Solution- Let ‘S’ be the sample space of throwing three dice.

∴ n(S) = 6³ = 216

Let ‘A’ be the event of getting a total of less than 6.

A = { (1, 1, 1) (1, 1, 2) (1, 2, 1) (2, 1, 1) (1, 1, 3) (1, 2, 2) (1, 3, 1) (2, 1, 2) (2, 2, 1) (3, 1, 1) }
n(A) = 10

P(A) = n(A)/n(S) = 10/216 = 5/108

∴ P(total of at least 6) = P(Ā)
⇒ P(total of at least 6) = 1 – P(A)
⇒ P(total of at least 6) = 1 – 5/108 = 103/108

Example 7: What is the probability that the number selected at random from the numbers 1, 2, 3, ……, 25 is a prime number?

Solution- 1 number out of 25 numbers can be selected in ²⁵C₁ ways.
∴ Total number of exhaustive cases = ²⁵C₁ = 25

Let ‘A’ be the event of getting a prime number.
∴ A = { 2, 3, 5, 7, 11, 13, 17, 19, 23 }

Out of these 9 prime numbers, one prime number can be selected in ⁹C₁ ways.
∴ Total number of favorable cases = ⁹C₁ = 9

∴ P(A) = number of favorable cases/number of exhaustive cases
⇒ P(A) = 9/25

Example 8: A ticket is drawn from 25 tickets numbered from 1, 2, 3,……, 25. Find the probability that the ticket drawn is a multiple of 3 or 7.

Solution- 1 ticket out of 25 tickets can be selected in ²⁵C₁ ways.
∴ Total number of exhaustive cases = ²⁵C₁ = 25

Let ‘A’ be the event of getting a ticket which is a multiple of 3 or 7.
∴ A = { 3, 6, 9, 12, 15, 18, 21, 24, 7, 14 }

Out of these 10 tickets, one ticket can be drawn in ¹⁰C₁ ways.
∴ Total number of favorable cases = ¹⁰C₁ = 10

∴ P(A) = number of favorable cases/number of exhaustive cases
⇒ P(A) = 10/25 = 2/5