Question

two dice are thrown at the same time. find the probability that the sum of the two numbers appearing on the top of the dice is more than 9​

Answer 1

Given :-

Two dice are thrown at the same time.

To Find :-

The probability that the sum of the two numbers appearing on the top is more than 9.

Solution :-

The sample space of the given event is given by,

$\sf (1,1) (1,2) (1,3)(1,4)(1,5)(1,6)$

$\sf (2,1)(2,2)(2,3)(2,4)(2,5)(2,6)$

$\sf (3,1)(3,2)(3,3)(3,4)(3,5)(3,6)$

$\sf (4,1)(4,2)(4,3)(4,4)(4,5)(4,6)$

$\sf (5,1)(5,2)(5,3)(5,4)(5,5)(5,6)$

$\sf (6,1)(6,2)(6,3)(6,4)(6,5)(6,6)$

According to the question,

Here, the total number of events = 36

By the formula,

$\underline{\boxed{\sf Probability=\dfrac{Number \ of \ favorable \ outcomes}{Number \ of \ outcomes} }}$

By substituting their values,

$\sf Probability=\dfrac{6}{36} =\dfrac{1}{6}$

Therefore, the probability is 1/6

To Note :-

Probability = Number of favorable outcomes/Total Number of outcomes

The probability is to find possibility of an event to happen which is equal to the ratio of the number of favorable outcomes and the total number of outcomes.

Answer 2

Given :-

• Two dice are thrown at the same time.

To Find :-

• The probability that the sum of the two numbers appearing on the top is more than 9.

Concept:-

✷The probability is to find possibility of an event to happen which is equal to the ratio of the number of favorable outcomes and the total number of outcomes.

$✷\underline{\boxed{\orange{\sf Probability=\dfrac{Number \ of \ favorable \ outcomes}{Number \ of \ outcomes} }} }$

SOLUTION:-

The sample space of the given event is :

$\sf (1,1) (1,2) (1,3)(1,4)(1,5)(1,6)(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)</p><p>\\ \\ </p><p>\sf (2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)</p><p>\\\\</p><p>\sf (3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)</p><p>\\\\</p><p>\sf (4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)</p><p>\\\\</p><p>\sf (5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)</p><p>\\\\</p><p>\sf (6,1)(6,2)(6,3)(6,4)(6,5)(6,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)$

Now we get to know that ,

the total number of events = 36

Then,

Substituting values in the formula,

$✧\purple{\sf Probability=\dfrac{Number \ of \ favorable \ outcomes}{Number \ of \ outcomes} }$

$\longrightarrow \sf Probability=\dfrac{6}{36} =\dfrac{1}{6}$

Hence, $\sf \boxed{\underline{\orange{the \: probability \: is \: 1/6}}}$

_______________________________________