Event: Sum on two dice' | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Probability |
In a throw of pair of dice, blue and grey, total no of possible outcomes=36(6×6) which 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)
E⟶ event of getting sum on 2 dice as 2
No. of favorable outcomes =1 (i.e.,(1,1))
Probability, P(E)=No.ofFavorableOutcomesTotalNo.ofPossibleOutcomes
P(E)=136
E⟶ event of getting sum as 3
No. of favorable outcomes =2(i.e.,(1,2)(2,1))
P(E)=236=118
E⟶ event of getting sum as 4
No. of favorable outcomes =3(i.e.,(3,1)(2,2)(1,3))
P(E)=336=112
E⟶ event of getting sum as 5
No. of favorable outcomes =4(i.e.,(1,4)(2,3)(3,2)(4,1))
P(E)=436=19
E⟶ event of getting sum as 6
No. of favorable outcomes =5(i.e.,(1,5)(2,4)(3,3)(4,2)(5,1))
P(E)=536
E⟶ event of getting sum as 7
No. of favorable outcomes =6(i.e.,(1,6)(2,5)(3,4)(4,3)(5,2)(6,1))
P(E)=636=16
E⟶ event of getting sum as 8
No. of favorable outcomes =5(i.e.,(2,6)(3,5)(4,4)(5,3)(6,2))
P(E)=536
E⟶ event of getting sum as 9
No. of favorable outcomes =4(i.e.,(3,6)(4,5)(5,4)(6,3))
P(E)=436=19
E⟶ event of getting sum as 10
No. of favorable outcomes =3(i.e.,(4,6)(5,5)(6,4))
P(E)=336=112
E⟶ event of getting sum as 11
No. of favorable outcomes =2(i.e.,(6,5)(5,6))
P(E)=236=118
E⟶ event of getting sum as 12
No. of favorable outcomes =1(i.e.,(6,6))
P(E)=136
From the figure(table) we can see that the outcomes are not equally likely - we see that, there is different probability for different outcome.
Hence cannot agree with the argument.