Two dice are thrown simultaneously find the probability of getting
1. an even number as sum
2. This sum as a prime number
3. A total of at least 10
4. A doublet of even number
5. A multiple of 2 on one dice and a multiple of 3 on the other
6. Same number on the both dice that is a doublet
7. A multiple of 3 as the sum
Answers
Given : Two dice are thrown simultaneously
To find : probability of getting
1. an even number as sum
2. This sum as a prime number
3. A total of at least 10
4. A doublet of even number
5. A multiple of 2 on one dice and a multiple of 3 on the other
6. Same number on the both dice that is a doublet
7. A multiple of 3 as the sum
Solution:
Dice 1- 6
Two dice are thrown
possible outputs = 6 * 6 = 36
an even number as sum if both Dice has odd numbers or both dice has even numbers
Probability = (3/6)(3/6) + (3/6)(3/6)
= 18/36
= 1/2
This sum as a prime number
2 , 3 , 5 , 7 , 11
2 = ( 1 , 1)
3 = ( 1, 2) , ( 2 , 1)
5 = ( 1 , 4) , ( 2 , 3) , ( 3 , 2) , ( 4 , 1)
7 = ( 1, 6) , ( 2 , 5) , ( 3 , 4 ) , ( 4 , 3) , ( 5 , 2) , ( 6 , 1)
11 = ( 5 , 6) , ( 6 , 5)
Probability = 15/36 = 5/12
A total of at least 10
10 , 11 or 12
10 = ( 4 , 6) , ( 5 , 5) , ( 6 , 4)
11 = ( 5 ,6 ) , ( 6 , 5)
12 = ( 6, 6)
Probability = 6/36 = 1/6
A doublet of even number
( 2,2) , (4 , 4) & (6,6)
Probability = 3/36 = 1/12
2 , 4 , 6 - multiple of 2
3 , 6 - multiple of 3
(2 , 3) , ( 2 , 6 ) , ( 3 , 2 , ) , (3 , 4) , ( 3 , 6) , ( 4 , 3) , ( 4 , 6) , ( 6 , 2) , (6 , 3) , ( 6 , 4) , ( 6 , 6)
Probability = 11/36
Doublet
(1,1) , ( 2,2) , (3,3), (4 , 4) , ( 5 ,5 ) & (6,6)
Probability =6/36 = 1/6
A multiple of 3 as the sum
3 , 6 , 9 , 12
3 = ( 1 , 2) ( 2 ,1 )
6 = ( 1 , 5) , ( 2 , 4) , (3 , 3) , (4 , 2) , ( 5 ,1 )
9 = (3 ,6) , ( 4 , 5) , ( 5 , 4) , ( 6 , 3)
12 = ( 6 , 6)
Probability = 12/36 = 1/3
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