emcrazy14 Group Title Two die are rolled. Find the probability that the numbers are the same given that at least one is 3? 2 years ago 2 years ago

1. cinar Group Title

|dw:1338882073720:dw|

2. cinar Group Title

can you expand your question, or write it in another way..

3. cinar Group Title

I did not get it..

4. emcrazy14 Group Title

I have to find the probability of getting two same numbers on both die if the number on one of them is atleast 3. I'm not sure myself. But I think it means that the number is either 3 or greater than 3.

5. cinar Group Title

|dw:1338882463299:dw|

6. cinar Group Title

6/36 prob. of the same number

7. cinar Group Title

|dw:1338882580009:dw|

8. cinar Group Title

at least one is 3 11/36

9. cinar Group Title

that's all I can do, sorry..

10. emcrazy14 Group Title

Are you sure this is the answer? What I'm thinking is that there are 11 outcomes in which one of the number is 3. But only 1 event in which 3,3 appears. So the probability should by 1/11. What do you think?

11. emcrazy14 Group Title

@experimentX : I didnot understand what you did. Would you please explain in another way?

12. emcrazy14 Group Title

So what would be the final answer?

13. kropot72 Group Title

The probability that at least one number is a 3 is$\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$ The probability that the other number will also be a 3 is 1/6 The overall probability of two 3's is therefore$\frac{1}{3}\times \frac{1}{6}=\frac{1}{18}$

14. emcrazy14 Group Title

Everyone has posted a different answer. Who should I trust? :(

15. experimentX Group Title

toss a coin ... lol

16. emcrazy14 Group Title

No, seriously. I'm more confused than ever before. :(

17. FoolForMath Group Title

Sample set in a role of two dice: $$\begin{array}{c|c|c|C|} &1&2&3&4&5&6 \\ \hline 1 &(1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\ \hline 2 &(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\ \hline 3 &(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\ \hline 4 &(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\ \hline 5 &(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\ \hline 6 &(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\\ \hline \end{array}$$

18. FoolForMath Group Title

Observe the table I posted above, You will see that that there are $$11$$ cases where either of the two dice show a $$3$$. And, among those $$11$$ cases only 1 i.e$$(3,3)$$ show both $$3$$. Ergo, the probability is $$\frac 1 {11} .$$

19. FoolForMath Group Title

You can also interpret this problem that we already know that one of the dice is 3 now the probability of another side being also 3 is $$\frac 1 6$$

20. FoolForMath Group Title

I can't see any blemish in either of these approaches, however I prefer #2. May be @Zarkon can see something else :)

21. kropot72 Group Title

If a pair of dice are rolled what is the probability that one die will roll a three?

22. FoolForMath Group Title

I think it's $$\frac {11}{36}$$

23. FoolForMath Group Title

assuming at-least, if it is strictly one then it is $$\large\frac{10}{36}$$

24. kropot72 Group Title

The Addition Theorem states that if an event can happen in a number of different and mutually exclusive ways, the probability of it happening at all is the sum of the separate probabilities that each event happens. The event is to roll a three. each die has a 1/6 probability of rolling a 3. therefore the probability that a 3 comes up on one or other of the dice is 1/6 + 1/6 = 2/6

25. kropot72 Group Title

@FoolForMath Do you intend to reply?

26. FoolForMath Group Title

They are rolled simultaneously not one after another. The sample space is 36 and 11 of them favor the even where atleast one die will roll a three.

27. kropot72 Group Title

The Addition Theorem holds. The simultaneous rolling does not affect the 1/6 probability of any of the 6 numbers coming up on either die.

28. Zarkon Group Title

"If a pair of dice are rolled what is the probability that one die will roll a three?" $P(3_1 \text{ or } 3_2)=P(3_1)+P(3_2)-P(3_1 \text{ and } 3_2)$ $=\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{11}{36}$ or $P(3_1 \text{ or } 3_2)=1-P(\left(3_1 \text{ or } 3_2\right)')=1-P(3'_1 \text{ and } 3'_2)$ $=1-\frac{5}{6}\times\frac{5}{6}=1-\frac{25}{36}=\frac{11}{36}$ the answer to the OP is $$\displaystyle\frac{1}{11}$$ E=equal , T=at least one 3 $P(E|T)=\frac{P(T|E)P(E)}{P(T)}=\frac{\frac{1}{6}\times\frac{1}{6}}{\frac{11}{36}}=\frac{1}{11}$

29. kropot72 Group Title

@Zarkon Thank you for your explanation. However it has a condition that only one die of the pair will have a three. The Addition Theorem holds where the probability of both dice showing 3 is included.

30. Zarkon Group Title

the condition, from the OP, is 'given that at least one is 3' not 'only one die of the pair will have a three'

31. kropot72 Group Title

@Zarkon Good point. Therefore the probability that at least one is a 3 when a pair of fair dice is rolled is 1/6 + 1/6 = 2/6

32. Zarkon Group Title

these events are not mutually exclusive. the way you are doing it you are double counting (3,3)

33. kropot72 Group Title

The number that comes up on die #1 has no bearing on the number that comes up on die #2. The sub-events 'number thrown on die #1' and 'number thrown on die #2' are indeed mutually exclusive. The sample space diagram has not bearing on this particular calculation.

34. Zarkon Group Title

you are describing independence ...the events are independent but they are not mutually exclusive.

35. vic1 Group Title

The probability of getting the same numbers = the probability of getting (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). Which is equals 6/36 =0.167 But, we are asked to fine the probability of getting the same numbers giving that at least one is 3 means the probability of getting 3 and above. Which the probability of getting (3,3), (4,4), (5,5), (6,6). Which equals 4/36 = 0.11

36. Zarkon Group Title

"we are asked to fine the probability of getting the same numbers giving that at least one is 3 means the probability of getting 3 and above" Really?

37. vic1 Group Title

@ Zarkon, I think it means at least one of the numbers is 3. What do u think?

38. Zarkon Group Title

"I think it means at least one of the numbers is 3." yes it doesn't mean "the probability of getting 3 and above"

39. kropot72 Group Title

@Zarkon Thank you very much for your patience and explanations. I accept that the Addition Theorem does not apply in this problem :)