Probability of Getting anything on Rolling or Throwing Two Dice

A die has six face marked 1, 2, 3, 4, 5 and 6. If we have more than one die, then all dice are considered as distinct, if not otherwise stated.

📌 Ex1. Two dice are thrown simultaneously. What is the probability that:
(i) 4 will not come up either time?
(ii) 4 will come up at least once?
✍ Solution:
When two dice are thrown, total possible outcomes =36
(i) Number of favorable outcomes that 4 will not come up either time is equal to (6-1)×(6-1)=25.

P(4 will not come up) =25/36

(ii) P(4 will come up once)=1-P(4 will not come up either time)
P(4 will come up once)=1-25/36=11/36

That means there are 11 favorable outcomes, they are
sample space e

📌 Ex2. The faces of a red cube and a yellow cube are numbered from 1 to 6. Both cubes are rolled. What is the probability that the top face of each cube will have the same number?
✍ Solution:
Total number of outcomes when both cubes are rolled =6×6=36 which are
E→ event of getting same number on each cube
Number of favorable outcomes n(E)=6 which are
{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}
P(E)=n(E)/n(S) =6/36=⅙

📌 Ex3. Two dice are thrown simultaneously. What is the probability that:
(i) 5 will not come up on either of them?
(ii) 5 will come up on at least one?
(iii) 5 wifi come up at both dice?
✍ Solution:
Total number of possible outcomes when 2 dice are thrown =6×6=36
(i) E→ event of 5 not coming up on either of them
Number of favorable outcomes =5×5=25. The probability is

P(E)=25/36

(ii) E→ event of 5 coming up at least once
sample space f

(iii) E→ event of getting 5 on both dice
Number of favorable outcomes =1; E={5,5}
P(E)=1/36

📌 Ex4. In a single throw of two dice, find the probability of:
(i) a doublet
(ii) a number less than 3 on each die
(iii) an odd number as a sum
(iv) a total of at most 10
(v) an odd number on one die and a number less than or equal to 4 on the other die.
✍ Solution:
The number of possible outcomes is 6×6=36. We write them as given below:

dice a

n(S)=36

(i) E= Event of getting a doublet
E={(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)};n(E)=6

Probability of getting a doublet
P=n(E)/n(S) =6/36=⅙

(ii) E= Event of getting a number less than 3 on each die
E={(1,1),(1,2),(2,1),(2,2)};n(E)=4

Probability of getting a number less than 3 on each die
P=n(E)/n(S) =4/36=1/9

(iii) Of course, in the sample space, there are a half number of odd sums as many as even sums. Hence, each probability of both is equally likely.
Probability of getting an odd number as sum
P=n(E)/n(S) =(½×n(S))/n(S) =½

(iv) E= Event of getting a total of at most 10.
dice b

Therefore total number of favorable ways =n(E)=33
Probability of getting a total of at most 10
P=n(E)/n(S) =33/36=11/12

(v) E= Event of getting an odd number on dice 1 and a number less than or equal to 4 on dice 2.
dice c

Expected pairs are shaded in yellow. Therefore, the number of expected pairs =n(E)=20
Probability of getting an odd number on dice 1 and a number less than or equal to 4 on dice 2
P=n(E)/n(S) =20/36=5/9

📌 Ex5. A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.
In throwing a dice, total possible outcomes =(1,2,3,4,5,6); n(S)=6.
For two dice, n(S)=6×6=36
Favorable cases where the sum is 10 or more with 5 on 1st die ={(5,5),(5,6)};n(E)=2.
Probability of getting a sum of 10 or more with Son 1st die

P=n(E)/n(S) =2/36=1/18

📌 Ex6. Two dice are rolled together. Find the probability of getting:
(i) a total of at least 10.
(ii) a multiple of 2 on one die and an odd number on the other die.
✍ Solution:
In throwing a dice, total possible outcomes ={1,2,3,4,5,6};n(S)=6.
For two dice, n(S)=6×6=36.
(i) E= event of getting a total of at least 10

={(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)};n(E)=6

Probability of getting a total of at least 10
P=n(E)/n(S) =6/36=⅙

(ii) E= event of getting a multiple of 2 on one die and an odd number on the other. It means that the sum of each expected pair is odd, because sum of an even and an odd is odd. Odd sum takes a half number of elements inside the sample space as many as even sum.
Probability of getting a multiple of 2 on one die and an odd number on the other
P=n(E)/n(S) =(½×36)/36=½

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