Question

two different dice are tossed together find the probability of doublet and of getting a sum of 10,the numbers on the two dice

Answer 1

As double dies are thrown therefore the total outcomes become 36.The outcomes are as following
(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).
1) Now to have a doublet we should found same numbers between brackets such as (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)
Therefore the favourable cases are 6
Probability =
$= \frac{6}{36} = \frac{1}{6}$
2)We should count outcomes having sum of 10 such that (4,6),(5,5)(6,4).So favourable outcomes are 3
Probability =
$= \frac{3}{36} = \frac{1}{12}$
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Answer 2

Given:

Two dice

To find:

The probability of getting a double

The probability of getting a sum of 10

Solution:

Sample space = ( 1, 0 ), ( 1, 1 ), ( 1, 2 )... ( 6, 6 )

Hence,  

Sample space = 36

To find the probability of getting a doublet,

Outcomes = ( 1, 1 ), ( 2, 2 ).. ( 6, 6 )

Possible outcomes = 6

P (getting a double) = 6 / 36  

Hence, P (getting a double) = 1 / 6

To find the probability of getting a sum of 10,

Possible outcomes = ( 4, 6 ), ( 6, 4 ), ( 5, 5 )  

P (getting a sum of 10 ) =3/36

Hence, P ( getting a sum of 10 ) = 1 / 12

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