Question

a die is rolled twice what is the probability that the sum is 9 or11?​

Answer 1

Given :- A die is rolled twice what is the probability that the sum is 9 or 11 ?

Solution :-

when a dice is rolled twice total Possible outcomes 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)
• Total 36.

and, total Possible outcomes of sum is 9 or 11 is :-

• (3,6) , (6,3) => sum 9.
• (5,4) , (4,5) => sum 9.
• (5,6) , (6,5) => sum 11.
• Total 6 .

then,

→ Required Probability = (total Possible outcomes of sum is 9 or 11) / (total Possible outcomes) = (6/36) = (1/6) (Ans.)

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Answer 2

SOLUTION

GIVEN

A die is rolled twice

TO DETERMINE

The probability that the sum is 9 or 11

EVALUATION

A die is rolled twice

So the total number of possible outcomes

$\sf{ = {6}^{2} }$

$= 36$

Let A be the event that the sum is 9

Then event points for the event A is (3,6),(4,5),(5,4),(6,3)

So the total number of possible outcomes for the event A is 4

So the probability of the event A

= P(A)

$= \displaystyle \sf{ \frac{4}{36} }$

Let B be the event that the sum is 11

Then event points for the event B is (5,6),(6,5)

So the total number of possible outcomes for the event B is 2

So the probability of the event B

= P(B)

$= \displaystyle \sf{ \frac{2}{36} }$

Now

A ∩ B is the event that the sum is 9 and 11

Now A and B are mutually exclusive

∴ A ∩ B = Φ

∴ P(A ∩ B) = 0

Hence the required probability that the sum is 9 or 11

= P(A ∪ B)

= P(A) + P(B) - P(A ∩ B)

$= \displaystyle \sf{ \frac{4}{36} + \frac{2}{36} - 0 }$

$= \displaystyle \sf{ \frac{6}{36} }$

$= \displaystyle \sf{ \frac{1}{6} }$

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