Two dice are thrown. what is the probability of getting a total of seven?

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 ` .

Probability means Possibility. It states how likely an event is about to happen. The probability of an event can exist only between 0 and 1 where 0 indicates that event is not going to happen i.e. Impossibility and 1 indicates that it is going to happen for sure i.e. Certainty. 

The higher or lesser the probability of an event, the more likely it is that the event will occur or not respectively. 

For example – An unbiased coin is tossed once. So the total number of outcomes can be 2 only i.e. either “heads” or “tails”. The probability of both outcomes is equal i.e. 50% or 1/2.
So, the probability of an event is Favorable outcomes/Total number of outcomes. It is denoted with the parenthesis i.e. P(Event).

P(Event) = N(Favorable Outcomes) / N (Total Outcomes)

Note: If the probability of occurring of an event A is 1/3 then the probability of not occurring of event A is 1-P(A) i.e. 1- (1/3) = 2/3

What is Sample Space?

All the possible outcomes of an event are called Sample spaces.

Examples-

1. A six faced dice is rolled once. So, total outcomes can be 6 and 
   Sample space will be [1, 2, 3, 4, 5, 6]
2. An unbiased coin is tossed, So, total outcomes can be 2 and 
   Sample space will be [Head, Tail]
3. If two dice are rolled together then total outcomes will be 36 and 
   Sample space will be [ (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) ]

Types of Events

Independent Events: If two events (A and B) are independent then their probability will be 
P(A and B) = P (A ∩ B) = P(A).P(B) i.e. P(A) * P(B)

Example: If two coins are flipped, then the chance of both being tails is 1/2 * 1/2 = 1/4

Mutually exclusive events:

• If event A and event B can’t occur simultaneously, then they are called mutually exclusive events.
• If two events are mutually exclusive, then the probability of both occurring is denoted as P (A ∩ B) and 
  P (A and B) = P (A ∩ B) = 0
• If two events are mutually exclusive, then the probability of either occurring is denoted as P (A ∪ B) P (A or B) = P (A ∪ B)                   = P (A) + P (B) − P (A ∩ B)                   = P (A) + P (B) − 0    

                 = P (A) + P (B)

Example: The chance of rolling a 2 or 3 on a six-faced die is P (2 or 3) = P (2) + P (3) = 1/6 + 1/6 = 1/3

Not Mutually exclusive events: If the events are not mutually exclusive then

P (A or B) = P (A ∪ B) = P (A) + P (B) − P (A and B)

What is Conditional Probability?

For the probability of some event A, the occurrence of some other event B is given. It is written as P (A ∣ B)

P (A ∣ B) = P (A ∩ B) / P (B)

Example- In a bag of 3 black balls and 2 yellow balls (5 balls in total), the probability of taking a black ball is 3/5, and to take a second ball, the probability of it being either a black ball or a yellow ball depends on the previously taken out ball. Since, if a black ball was taken, then the probability of picking a black ball again would be 1/4, since only 2 black and 2 yellow balls would have been remaining, if a yellow ball was taken previously, the probability of taking a black ball will be 3/4.

Solution:

When two dice are rolled together then total outcomes are 36 and Sample space is [ (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) ]

So, pairs with sum 7 are (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1) i.e. total 6 pairs

Total outcomes = 36
Favorable outcomes = 6

Probability of getting the sum of 7 = Favorable outcomes / Total outcomes

                                                      = 6 / 36 = 1/6

So, P(sum of 7) = 1/6.

Similar Questions

Question 1: What is the probability of getting 1 on both dice?

Solution:

When two dice are rolled together then total outcomes are 36 and Sample space is [ (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) ]

So, pairs with both 1’s are (1,1) i.e. only 1 pair

Total outcomes = 36
Favorable outcomes = 6

Probability of getting pair of 1 = Favorable outcomes / Total outcomes 
                                                 = 1 / 36 

So, P(1,1) = 1/36.

Question 2: What is the probability of getting the sum of 4?

Solution:

When two dice are rolled together then total outcomes are 36 and Sample space is 

[ (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) ]

So, pairs with sum 4 are (1,3) (2,2) (3,1)  i.e. total 3 pairs

Total outcomes = 36
Favorable outcomes = 3

Probability of getting the sum of 4 = Favorable outcomes / Total outcomes
                                                       = 3/36 = 1/12

So, P(sum of 3) = 1/12.

Question 3: What is the probability of getting the sum of 5?

Solution:

When two dice are rolled together then total outcomes are 36 and Sample space is [ (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) ]

So, pairs with sum 5 are (1,4) (2,3) (3,2) (4,1) i.e. total 4 pairs

Total outcomes = 36
Favorable outcomes = 4

Probability of getting the sum of 5 = Favorable outcomes / Total outcomes
                                                       = 4 / 36 = 1/9

So, P(5) = 1/9.

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