# Probability of Getting anything on Rolling or Throwing a Die

## Throwing a die once

Ex1. A die is rolled, find the probability that an even number is obtained.
Solution:
Let us first write the sample space S of the experiment.

S={1,2,3,4,5,6}→n(S)=6

Let E be the event “an even number is obtained” and write it down.
E={2,4,6}→n(E)=3

We now use the formula of the classical probability.
P(E)=n(E)/n(S)=3/6=½

Ex2. A die is rolled, find the probability that the number obtained is greater than 4.
Solution:

S={1,2,3,4,5,6};E={5,6};n(S)=6;n(E)=2
P(E)=n(E)/n(S) =2/6=⅓

Ex3. A die is thrown once. Find the probability of getting
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
solutions:
The possible outcomes when a dice is thrown {1,2,3,4,5,6}
Number of possible outcomes of a dice =6
(i) Prime numbers on a dice are 2, 3, and 5.
Total prime numbers on a dice =3
Probability of getting a prime number =3/6=½
(ii) Numbers lying between 2 and 6 = 3, 4, 5
Total numbers lying between 2 and 6 =3
Probability of getting a number lying between 2 and 6 =3/6=½
(iii) Odd numbers on a dice = 1, 3, and 5
Total odd numbers on a dice =3
Probability of getting an odd number =3/6=½

Ex4. In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will not be an even number.
(iii) will be an odd number.
solutions:
Sample space ={1,2,3,4,5,6}
n(S)=6
(i) E= event of getting an even number ={2,4,6}

n(E)=3

Probability of a getting an even number
=n(E)/n(S) =3/6=½

(ii) E= event of not getting an even number ={1,3,5}
n(E)=3

Probability of a not getting an even number
=n(E)/n(S) =3/6=½

(iii) E= event of getting an odd number ={1,3,5}
n(E)=3

Probability of a getting an odd number
=n(E)/n(S) =3/6=½

Ex5. In a single throw of die, find the probability of getting:
Sampie space ={1,2,3,4,5,6}

n(S)=6

(i) E= event of getting a 5 on a throw of die ={5}
n(E)=1

Probability of getting a 5
P(E)=n(E)/n(S) =⅙

(ii) There are only six possible outcomes in a single throw of a die. If we want to find probability of 8 to come up, then in that case number of possible or favourable outcome is 0 (zero)
n(E)=0

Probability of getting a 8
P(E)=n(E)/n(S) =0/6=0

(iii) If we consider to find the probability of number less than 8. then all six cases are favourable n(E)=6
Probability of getting a number less than 8
P(E)=n(E)/n(S) =6/6=1

(iv) E= event of getting a prime number ={2,3,5}
n(E)=3

Probability of getting a prime number
P(E)=n(E)/n(S) =3/6=½

Ex6. A die is thrown once. Find the probability of getting:
(i) an even number
(ii) a number between 3 and 8
(iii) an even number or a multiple of 3
solutions:
Sample space ={1,2,3,4,5,6}

n(S)=6

(i) E= the possible even numbers ={2,4,6}
n(E)=3

Probability of getting an even number
P(E)=n(E)/n(S) =3/6=½

(ii) E= the possible even numbers between 3 and 8 ={4,5,6}
n(E)=3

Probability of getting an even number between 3 and 8
P(E)=n(E)/n(S) =3/6=½

(iii) E= the event of getting an even number or a multiple of 3 ={2,3,4,6}
n(E)=4

Probability of getting an even number or a multiple of 3
P(E)=n(E)/n(S) =4/6=⅔

Ex7. In a single throw of a die, find the probability of getting a number:
(i) greater than 4.
(ii) less than or equal to 4.
(iii) not greater than 4.
solutions:
Sample space ={1,2,3,4,5,6}
n(S)=6
(i) E= event of getting a number greater than 4 ={5,6}

n(E)=2

Probability of a number greater than 4
=n(E)/n(S) =2/6=⅓

(ii) E= event of getting a number less than or equal to 4 ={1,2,3,4}
n(E)=4

Probability of a number less than or equal to 4
=n(E)/n(S) =4/6=⅔

(iii) E= event of getting a number not greater than 4 ={1,2,3,4}
n(E)=4

Probability of a number not greater than 4
=n(E)/n(S) =4/6=⅔

Ex8. A die is thrown once. Find the probability of getting a number:
(i) less than 3
(ii) greater than or equal to 4
(iii) less than 8
(iv) greater than 6
solutions:
In throwing a dice, total possible outcomes ={1,2,3,4,5,6}

n(S)=6

(i) On a dice, numbers less than 3, expected elements ={1,2}, n(E)=2.
Probability of getting a number less than 3=
=n(E)/n(S) =2/6=⅓

(ii) On a dice, numbers greater than or equal to 4, expected elements ={4,5,6}, n(E)=3.
Probability of getting a number greater than or equal to 4
=n(E)/n(S) =3/6=½

(iii) On a dice, numbers less than 8, expected elements ={1,2,3,4,5,6}, n(E)=6
Probability of getting a number less than 8
=n(E)/n(S) =6/6=1

(iv) On a dice, numbers greater than 6, expected elements ={┤}=Ø, n(E)=0
Probability of getting a number greater than 6
=n(E)/n(S) =0/6=0

Ex9. A dice is thrown once. What is the probability of getting a number:
(i) greater than 2?
(ii) less than or equal to 2?
solutions:
Number of possible outcomes when dice is thrown

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

(i) Event of getting a number greater than 2 =E={3,4,5,6};n(E)=4
Probability of getting a number greater than 2
P=n(E)/n(S) =4/6=⅔

(ii) Event of getting a number less than or equal to 2 =E={1,2};n(E)=2
Probability of getting a number less than or equal to 2
P=n(E)/n(S) =2/6=⅓

Ex10. 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) an even prime number
(v) a number greater than 5
(vi) a number lying between 2 and 6
Solutions:
(i) Total number of possible outcomes =6, they are {1, 2, 3, 4, 5, 6}
E→ Event of getting a prime number.
Number of favorable outcomes =3, they are {2,3,5} (ii) E→ Event of getting 2 or 4.
Number of favorable outcomes =2, they are {2,4}
Total number of possible outcomes =6
Then, P(E)=2/6=⅓
(iii) E→ Event of getting a multiple of 2 or 3
Number of favorable outcomes =4, they are {2,3,4,6}
Total number of possible outcomes =6
Then, P(E)=4/6=⅔
(iv) E→ Event of getting an even prime number
Number of favorable outcomes =1, {2}
Total number of possible outcomes =6, they are {1,2,3,4,5,6}
P(E)=⅙

(v) E→ Event of getting a number greater than 5.
Number of favorable outcomes =1, {6}
P(E)=⅙

(vi) E→ Event of getting a number lying between 2 and 6.
Number of favorable outcomes =3, they are {3,4,5}
P(E)=3/6=½

Ex11. A die is thrown, find the probability of following events:
(i). A prime number will appear,
(ii). A number greater than or equal to 3 will appear,
(iii). A number less than or equal to one will appear,
(iv). A number more than 6 will appear,
(v). A number less than 6 will appear.
Solutions:
The sample space of the given experiment is given by

S={1,2,3,4,5,6}

(i). Let A be the event of the occurrence of a prime number. Accordingly, A={2,3,5}
P(A)=n(A)/n(S) →P(A)=3/6=½

(ii). Let B be the event of the occurrence of a number greater than or equal to 3. Accordingly, B={3,4,5,6}
P(B)=n(B)/n(S) →P(B)=4/6=⅔

(iii). Let C be the event of the occurrence of a number less than or equal to one. Accordingly, C={1}
P(C)=n(C)/n(S) →P(C)=⅙

(iv). Let D be the event of the occurrence of a number greater than 6. Accordingly, D=Ø
P(D)=n(D)/n(S) →P(D)=0/6=0

(v) Let E be the event of the occurrence of a number less than 6. Accordingly, E={1,2,3,4,5}
P(E)=n(E)/n(S) →P(E)=⅚

Ex12. A die is rolled. Find the probability in each case: The top shows events-happening:
(i). 3 or 4 dots
Solution:
Sample space ={1,2,3,4,5,6}
Total outcomes =n(S)=6

P(3 or 4 dots)=P(3)+P(4)=⅙+⅙=⅓

(ii) Dots less than 5
Solution:
Sample space ={1,2,3,4,5,6}
Total outcomes =n(S)=6
P{dots less than 5)=P(1 or 2 or 3 or 4)
=P(1)+P(2)+P(3)+P(4)=⅙+⅙+⅙+⅙=⅔

## Throwing a die twice

Ex13. A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment].
solutions:
Total number of outcomes =6×6=36
(i) Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)
Hence, total number of favorable cases =11
P(5 will come up either time)=11/36
(ii) Total number of cases, when 5 can come at least once =11
P(5 will come at least once)=11/36

Ex14. A die is numbered in such a way that its faces show the number 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws: What is the probability that the total score is (i) even? (ii) 6? (iii) at least 6?
Solutions: Total number of possible outcomes when two dice are thrown =6×6=36
(i) Total times when the sum is even =18
P(getting an even number)=18/36=½
(ii) Total times when the sum is 6 =4
P(getting sum as 6)=4/36=1/9
(iii) Total times when the sum is at least 6 (i.e., greater than 5) =15
P(getting sum at least 6)=15/36=5/12

## Rolling a die and tossing a coin at the same tine

Ex15. A die is rolled and a coin is tossed, find the probability that the die shows an odd number and the coin shows a head.
Solution:
The sample space S of the experiment described in question 5 is as follows Let E be the event “the die shows an odd number and the coin shows a head”. Event E may be described as follows
E={(1,H),(3,H),(5,H)};n(E)=3

The probability P(E) is given by
P(E)=n(E)/n(S) =3/12=¼

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