When two dice are thrown, the probability of getting 10 or 11 is

Solution:

When two dice are thrown simultaneously, the sample space of the experiment 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 there are 36 equally likely outcomes.

Possible number of outcomes = 36.

(i)Let E be an event of getting a doublet.

Favourable outcomes = {(1,1), (2,2),(3,3), (4,4), (5,5),(6,6)}

Number of favourable outcomes = 6

P(E) = 6/36 = 1/6

Probability of getting a doublet is 1/6 .

(ii)Let E be an event of getting a sum of 8.

Favourable outcomes = {(2,6), (3,5), (4,4), (5,3), (6,2)}

Number of favourable outcomes = 5

P(E) = 5/36

Probability of getting a sum of 8 is 5/36.

Hint: In this question, we are given that two dice are thrown simultaneously and we have to find various probabilities of numbers shown on both dice. For this, we will first make sample space and then use that to find favorable outcomes for finding each probability. Total outcomes will be given as the number of elements in sample space. Probability of any event is given as $\text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{Total outcomes}}$.

Complete step by step answer:

Hence, the probability of getting the sum as 8 is $\dfrac{5}{36}$.

Hence, probability of getting a multiple of 2 on one dice and multiple of 3 on other dice is $\dfrac{11}{36}$.

Hence, the probability of getting a sum at least 10 is $\dfrac{1}{6}$.

Note: Students should carefully count all the possibilities while calculating probability. In (ii) part, students should note that multiple of 2 or 3 can be on any of the two dices. For example, (2,3) and (3,2) both are favorable cases. In (iii) part, students should note that sum should be at least 10, so, they have to consider sum as 10 or higher. Try to avoid mistakes while making sample space.

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-

• A six-faced dice is rolled once. So, total outcomes can be 6 and 
  Sample space will be [1, 2, 3, 4, 5, 6]
• An unbiased coin is tossed, So, total outcomes can be 2 and 
  Sample space will be [Head, Tail]
• 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 
   (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 10 are (4,6) (5, 5) (6, 4) i.e. total 3 pairs

Total outcomes = 36
Favorable outcomes = 3

Probability of getting the sum of 10 = Favorable outcomes / Total outcomes

                                                        = 3 / 36 = 1/12

So, P(sum of 10) = 1/12

Similar Questions

Question 1: What is the probability of getting a sum of 11 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 sum 11 are (5,6) (6,5) i.e. only 2 pairs

Total outcomes = 36
Favorable outcomes = 2

Probability of getting pair with sum 11 = Favorable outcomes / Total outcomes
                                                              = 2 / 36 = 1/18

So, P(sum of 11) = 1/18.

Question 2: What is the probability of getting the sum of 12?

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 12 are (6,6)  i.e. only 1 pair

Total outcomes = 36
Favorable outcomes = 1

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

So, P(sum of 12) = 1/36.

Question 3: What is the probability of getting the sum of 9 with two 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 sum 9 are (3,6) (4,5) (5,4) (6,3) i.e. total 4 pairs

Total outcomes = 36
Favorable outcomes = 4

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

So, P(9) = 1/9.

Page 2

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-

• A six-faced dice is rolled once. So, total outcomes can be 6 and 
  Sample space will be [1, 2, 3, 4, 5, 6]
• An unbiased coin is tossed, So, total outcomes can be 2 and 
  Sample space will be [Head, Tail]
• 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 14 are 0.

Total outcomes = 36
Favorable outcomes = 0

Probability of getting the sum of 14 = Favorable outcomes / Total outcomes

                                                        = 0/36 = 0

So, P(sum of 14) = 0

Similar Questions

Question 1: What is the probability of getting the sum of 11 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 sum 11 are (5,6) and (6,5) i.e. only 2 pairs

Total outcomes = 36
Favorable outcomes = 2

Probability of getting the sum of 11 = Favorable outcomes / Total outcomes 
                                                         = 2 / 36 = 1/18

So, P(sum of 11) = 1/18.

Question 2: What is the probability of getting the product of 6 on two 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 product 6 are (1,6) (2,3) (3,2) (6,1)  i.e. total 4 pairs

Total outcomes = 36
Favorable outcomes = 4

Probability of getting the product of 6 = Favorable outcomes / Total outcomes                                                                                                                         = 4/36 = 1/9

So, P(product of 6) = 1/9.

Question 3: What is the probability of getting a pair with an even sum on two 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) ]

Possible even sums are 2, 4, 6, 8, 10, 12

So, possible pairs with sum 2, 4, 6, 8, 10 and 12 are (1,1) (2,2) (1,3) (3,1) (1,5) (2,4) (4,2) (3,3) (5,1) (2,6) (3,5) (4,4) (5,3) (6,2) (4,6) (5,5) (6,4) (6,6) i.e. total 18 pairs

Total outcomes = 36
Favorable outcomes = 18

Probability of getting the sum of 5 = Favorable outcomes / Total outcomes 
                                                       = 18 / 36 = 1/2

So, P(Even sum) = 1/2.