(1
EXAMPLE 20 Two dice are thrown simultaneously. Find the probability of getting
an even number as the sum
(ii) the sum as a prime muumba
(iii) a total of at least 10
(iv) a doublet of een member
(v) a multiple of 2 on one dice and a multiple of 3 on the other.
(vi) same number on both dice i.e. a doublet.
(CBSE
(vii) a multinle of 3 as the sum.
Answers
Answer:
- 5/12
- 1/6
- 1/13
- 11/36
- 1/3
Step-by-step explanation:
Sample space for total number of possible outcomes
(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),
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
Total number of outcomes =36
(i)
Favorable outcomes for sum as prime are
(1,1),(1,2),(1,4),(1,6),(2,3),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)
Number of favorable outcomes =15
Hence, the probability of getting the sum as a prime number. = 15/3 = 5/12
(ii)
Favorable outcomes for total of atleast 10 are
(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)
Number of favorable outcomes =6
Hence, the probability of getting a total of atleast 10 = 6/36 = 1/6
(iii)
Favorable outcomes for a doublet of even number are
(2,2),(4,4),(6,6)
Number of favorable outcomes =3
Hence, the probability of getting a doublet of even number = 3/36 = 1/13
(iv)
Favorable outcomes for a multiple of 2 on one dice and a multiple of 3 on the other dice are
(2,3),(2,6),(3,2),(3,4),(3,6),(4,3),(4,6),(6,2),(6,3),(6,4),(6,6)
Number of favorable outcomes =11
Hence, the probability of getting a multiple of 2 on one dice and a multiple of 3 on the other dice =11/36
(v)
Favorable outcomes for getting a multiple of 3 as the sum
(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3)(6,6)
Number of favorable outcomes =12
Hence, the probability of getting a multiple of 3 as the sum = 12/36 = 1/3
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Given : Two dice are thrown simultaneously
To find : probability of getting
an even number as the sum
(ii) the sum as a prime muumba
(iii) a total of at least 10
(iv) a doublet of een member
(v) a multiple of 2 on one dice and a multiple of 3 on the other.
(vi) same number on both dice i.e. a doublet.
(vii) a multinle 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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