Question

Two diflerent dice are thrown at the same time. Find the probability of getting :
(i) the sum of numbers on the two dice, greater than 9.
(ii) the surn of numbers on the two dice, less than 6.

Answer 1

 Two dice tossed simultaneously , than total number of events ( outcomes ) =  6 × 6 = 36

As we can show  all the 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 }


i)

Possible outcomes =

{ 4,6 },{ 5,5 } , { 5,6 }, { 6, 4 },{ 6 ,5 } , { 6,6 }


No. Of possible outcomes = 6


Probability (P) = 6/36= ⅙


ii)

Possible outcomes=  {1,1 } , {1,2 } {1,3 }, {1,4 }

{ 2,1 } , {2 ,2 } , {2,3 },{ 3,1 } ,{ 3,2 },{ 4,1 }


No. Of possible outcomes = 10



Probability (P) = 10/36= 5/18



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Hope this will help you....

Answer 2

Both Dice tossed simultaneously, so the total number of events = 6×6 == 36....

So, the all possible Outcomes will be ----

(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..
1°°°° possible outcomes== (4,6):(5,5):(5,6):(6,4):(6,5):(6,6)...
No. of thier outcomes is 6
SO, PROBABILITY (p)→6/36→1/6 ..

2°°°°In this Possible outcomes---> (1,1):(1,2):(1,3):(1,4)
(2,1):(2,2):(2,3):(3,1):(3,2):(4,1)

So In this No. of possible outcomes-----10 ..

Hence, Probability (P) = 10/36== 5/18...