Two dice are thrown once the probability of getting a sum of at least 9 is a 0 b 1 36 c 5/18 d 7 36

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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:

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.

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.

What is the probability that sum on both faces is 9 when two dice are thrown simultaneously?

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(sum of 9) = 1/9.

Similar Questions

Question 1: What is the probability of getting a sum of 10 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 sum 10 are (4,6) (5,5) (6,4) i.e. only 3 pairs

Total outcomes = 36
Favorable outcomes = 3

Probability of getting sum 10 = Favorable outcomes / Total outcomes 
                                              = 3 / 36 = 1/12

So, P(10) = 1/12.

Question 2: What is the probability of getting the sum of 11?

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. total 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 3: 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(12) = 1/36