Two dice are thrown, write all the
possible outcomes and find the
probability of getting:
multiple of 4
as the sum
Answers
Answer:
The possible outcomes when two dice are thrown together are:
(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).
Therefore, the number of possible outcomes when two dice are thrown is 36.
Step-by-step explanation:
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Answer:
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. =
36
15
=
12
5
(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 =
36
6
=
6
1
(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 =
36
3
=
13
1
(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 =
36
11
(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 =
36
12
=
3
1