Two dice are thrown together what is the probability of getting a sum of 7

Probability = possible outcomes/
total outcomes

when a single die is thrown the number of outcomes is 6^1

when two dice are thrown the number of outcomes is 6^2

In throwing two dice the favorable cases of getting the sum as 7 are: (1, 6), (6, 1),
(2, 5), (5, 2), (3, 4), (4, 3).

Therefore,
the required probability is 6/36 or 1/6.

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The dice quantity = 2(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)Probability= 1+6=7 2+5=7 3+4=7 4+3=7 5+2=7 6+1=7So,answer=6/36

1/6

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Text Solution

Solution : The outcomes of the rolling of a pair of dice can be more readily understood by assigning <br> colors to the dice (e.g ., one red , one green ). An outcome of throwing aa pair of dice can be <br> represented by an ordered pair , where the first element is the side of the red die fecing up, <br> and the second is the side of the green die fecing up . Thus , (32) signifies a 3 on the red die <br> and a 2 on the green, while (23) signifies 2 on the red 3 on the green . Since there are ltbr. six feces on each die , there are ` 6xx6 = 36` possible outcomes in the same space . the <br> outcomes that have a sum of 7 are (61). (16), (52) , (25) , (43) and (34) . Therefore , the <br> probability of getting a sum of 7 is `(6)/(36) = (1)/(6)` <br> Probability questions can also be asked about two events A and B . Questions on the Math <br> Leval 2 Test cover the following three ways of combining two events <br> `*P(A cupB)` , the probability of either A or B occurring . where ` A cup B ` is called the disjunction <br> of A and B <br> ` * (PA|B)` , the probability of A occurring , givne that B has occurred . This is called conditional <br> Probability <br> ` * P(A nn B)` , the probabilty of both A and B occurring , where ` A nn B ` is called the conjunction <br> of A and B <br> If two event have no outcomes in common , they are called mutually exclusive , and `P(A uu B) = ` <br> P(A) + P(B) . If two event no outcomes in common , these outcomes would be double counted , <br> resulting i the more general formaula ` P(AuuB)= P(A) + P(A) - P(AnnB) ` . Note that A and B are <br> mutually exclusive if and only if ` P(A nn B ) = 0 ` .

Answer (Detailed Solution Below)

Option 4 : 1/6

Concept Used: 

P(A) = Favorable outcomes/Total number of outcomes

Calculations:

Total number of possible outcomes = 6 × 6 = 36

Favorable outcomes = (1, 6) (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

⇒ P(A) = 6/36 = 1/6

∴ The probability of getting sum as 7 when two dice are thrown is 1/6.

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Solution:

Sample space of rolling of 2 dice is as below,

S = {(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)}

n(S) = 36

Let A = sum of numbers is 7 = { (1, 6)(2, 5)(3, 4) (4, 3) (5, 2) (6, 1)}

n(A) = 6

P (Sum of numbers is 7) = n(A) / n(S)

= 6/36 = 1/6

Therefore, the probability of rolling two dice and getting a sum of 7 is 1/6.

Summary:

The probability of rolling two dice and getting a sum of 7 is 1/6.