Question

two dice are thrown simulataneously . then then write down the possible outcome dot and also find the probability of that outcomes in which the sum of outcome is odd​

Answer 1

Answer :-

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

Possible outcomes whose sum are odd -

➩ 1 , 2

➩ 1 , 4

➩ 1 , 6

➩ 2 , 1

➩ 2 , 3

➩ 2 , 5

➩ 3 , 2

➩ 3 , 4

➩ 3 , 6

➩ 4 , 1

➩ 4 , 3

➩ 4 , 5

➩ 5 , 2

➩ 5 , 4

➩ 5 , 6

➩ 6 , 1

➩ 6 , 3

➩ 6 , 5

Total outcomes = 18

Possibility = $\sf \frac{18}{36}$

$\sf = \frac{1}{2}$

$\boxed{\rm Possibility = \frac{1}{2}}$

Answer 2

Possible outcomes whose sum are odd -

( 1 , 2 )

( 1 , 4 )

( 1 , 6 )

( 2 , 1 )

( 2 , 3 )

( 2 , 5 )

( 3 , 2 )

( 3 , 4 )

( 3 , 6 )

( 4 , 1 )

( 4 , 3 )

( 4 , 5 )

( 5 , 2 )

( 5 , 4 )

( 5 , 6 )

( 6 , 1 )

( 6 , 3 )

( 6 , 5 )

Total outcomes = 18

$Probability = \frac{18}{36}$

$\bold {Probability = \frac{1}{2}}$