two dice(one blue and one grey) are thrown at the same time, write down all the possible outcomes.What is the probability that the sum of two nos. appearing on the top of the dice is: (i) 8  (ii) 13 (iii) less than or equal to 12 ?

Dear Student,

Please find below the solution to the asked query:

When we tossed a die , than total number of events ( outcomes ) = 6  , As : {  1 , 2 , 3 , 4 , 5 , 6 }

But when we tossed two dice simultaneously , than total number of events ( outcomes ) =  6 $\times$ 6 = 36 

As we can show  all the possible outcomes , As : { 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 }

So,

n ( S  ) = 36

We know probability P ( E )  =   $\frac{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{desired} \mathrm{events} \mathrm{n} ( \mathrm{E} )}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{events} \mathrm{n} ( \mathrm{S} )}$

i ) The probability that the sum of two numbers appearing on the top of the dice is  8

So,

Number of events = { 2 , 6 }, { 3 , 5 }, { 4 , 4 }, { 5 , 3 } ,{ 6 , 2 }

So,

n ( E ) =  5
The probability that the sum of two numbers appearing on the top of the dice is  8 = $\frac{\mathbf{5}}{\mathbf{36}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$

ii ) The probability that the sum of two numbers appearing on the top of the dice is  13

So,

Number of events = 0

So,

n ( E ) =  0
The probability that the sum of two numbers appearing on the top of the dice is 13= $\frac{0}{36}= \mathbf{0}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$

iii ) The probability that the sum of two numbers appearing on the top of the dice is  less than or equal to 12

So,

Number of events = { 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 }

So,

n ( E ) =  36
The probability that the sum of two numbers appearing on the top of the dice is less than or equal to 12 = $\frac{36}{36} = \mathbf{1}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$

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