Question

Find the probability of getting the sum of two numbers, less than 3 or more than 11, when a pair of distinct dice is thrown together?

Answer 1

When two dice are thrown, the possible outcomes or Sample Space = {(1,1), (1,2), (1,3), ..., (1,6), (2,1),(2,2),...,(2,6), (3,1),(3,2),...,(3,6), (4,1),....(4,6),(5,1)...,(5,6),(6,1),...(6,6)}

So there are total 36 outcomes whech two dice are thrown.

According to question, we have to find the sum of numbers such that they are less than 3 and more than 11.

So, Let E denote the number of outcomes for the required circumstance.

∴ E = {(1,1),(6,6)}

Thus, number of outcomes, n(E) = 2

∴ Required Probability = n(E)/S

= 2/36

= 1/18

Hence, 1/18 is the required probability.

Answer 2

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• Two distinct pair of dice is thrown.

$\LARGE{\bf{\underline{\underline{TO \ FIND:-}}}}$

• We need to find the probability when the:

          ⇒ Sum of the outcomes of the pair of dice is less than 3.

          ⇒ Sum of the outcomes of the pair of dice is more than 11.

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

CASE : 1-

When the sum of the outcomes of the pair of dice is less than 3.

There is only one case when the sum of outcomes is less than 3. If both dice gives "1" as the outcome, then the sum would be 1+1 = 2, which is less than 3. So, CASE 1 : ( 1 , 1 )

CASE : 2-

When the sum of the outcomes of the pair of dice is more than 11.

There is only one case when the sum of outcomes is more than 11. If both dice gives "6" as the outcome, then the sum would be 6 + 6 = 12, which is more than 11. So, CASE 2 : ( 6 , 6 )

So, we only have 2 events where the condition satisfies.

And total number of events (combinations) of two dices would be 6×6 = 36.

$\boxed{\sf{\red{P_{Req}=\dfrac{Number \ of \ favourable \ events}{Total \ number \ of \ events} }}}$

$\to \sf P_{req}=\dfrac{2}{36}$

$\to \sf{\red{P_{req}=\dfrac{1}{18}}}$

The required probability is 1/18.

Regards,

SujalSirmilla

Ex - brainly star.