What is the probability that their sum is 9?

`7/18``7/24``7/36``7/42`

Solution : Total number of outcomes in a throw of two dice `=36`
Favourable ways of getting a sum of `9=(3,6),(4,5),(5,4),(6,3)`
`implies` No. of favourable ways of getting a sum `9=4`
and the probability of getting `9=4/36`
Again, favourable ways of getting a sum of `10=(4,6),(5,5),(6,4)`
`implies` No. of favourable ways of getting a sum of `10=3`
and the probability of getting `10=3/36`
These two events are mutually exclusive.
`:.` Probability of getting a sum of 9 or 10.
`=4/36+3/36=7/36`

In problems like this (with small numbers and just two independent variables), I find it often helps to draw a table and mark the cases being looked for.

What is the probability that their product is a multiple of 3?

The product would be a multiple of 3 if either number is 3. I've marked those cases with the xs in this pretty little ASCII art table:

  123456789
1 ..x..x..x
2 ..x..x..x
3 xxXxxXxxX
4 ..x..x..x
5 ..x..x..x
6 xxXxxXxxX
7 ..x..x..x
8 ..x..x..x
9 xxXxxXxxX

You can eyeball count that as 6 lines of 9 xs, except that would double-count the ones where the lines intersect (uppercase Xs), so reduce 9 for that. Of course you could count them in some other way to verify.

That gives $$ P(\mathrm{product\ is\ multiple\ of\ 3}) = \frac{6 \cdot 9 - 9}{9 \cdot 9} = \frac{5 \cdot 9}{9 \cdot 9} = \frac{45}{81} = \frac{5}{9} $$

I used this logic: if any one of the digits selected is 3, then the product would be a multiple of 3. There are 3 multiples of 3 (3, 6, 9) between 1-9, so 3/9 = 1/3, and since the second number can be anything [...]

That first part is right. Just that I think in the follow-up you missed that the first number can also be anything if the second satisfies the condition. Listing the pairs explicitly helps with that, and the table also exposes the overlapping cases.

Practice problem: What's the probability if we change the problem so that the two digits can't be the same? (It's not the same, by my count.)

Statistics

1 Answer

Daniel L.

Jan 11, 2017

The probability is#1/9#. See explanation.

Explanation:

The sum of#9#can be rolled in#4#ways:

#A={(3,6),(4,5),(5,4),(6,3)}#

The total number of possible results is:#bar(bar(Omega))=6^2=36#

The probability is:

#P(A)=bar(bar(A))/bar(bar(Omega))=4/36=1/9#

Answer link

Related questions

• How do I determine the molecular shape of a molecule?

• What is the lewis structure for co2?

• What is the lewis structure for hcn?

• How is vsepr used to classify molecules?

• What are the units used for the ideal gas law?

• How does Charle's law relate to breathing?

• What is the ideal gas law constant?

• How do you calculate the ideal gas law constant?

• How do you find density in the ideal gas law?

• Does ideal gas law apply to liquids?

Impact of this question

38687 views around the world

You can reuse this answer
Creative Commons License

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 12 = Favorable outcomes / Total outcomes                                                                                                                         = 1 / 36

What is the probability of getting a sum of more than 9?

The probability of scoring a sum greater than 9 or a doublet is =10/36=5/18.

What is the probability of getting a sum 9 from two throws of a disc?

So, P(sum of 9) = 1/9.

What is the probability of getting a 9 or 10?

∴ The probability of getting a 9 or 10 on single throw of 2 dice is 7/36.

What is the probability of getting a sum as 9 if a dice is thrown thrice *?

Probability of a sum of 9: 25/216 = 11.6% Probability of a sum of 10: 27/216 = 12.5%