Two die are rolled. Find the probability that the numbers are the same given that at least one is 3?

- anonymous

Two die are rolled. Find the probability that the numbers are the same given that at least one is 3?

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- schrodinger

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- anonymous

|dw:1338882073720:dw|

- anonymous

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

- anonymous

I did not get it..

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## More answers

- anonymous

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.

- anonymous

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- anonymous

6/36 prob. of the same number

- anonymous

|dw:1338882580009:dw|

- anonymous

at least one is 3
11/36

- anonymous

that's all I can do, sorry..

- anonymous

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?

- anonymous

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

- anonymous

So what would be the final answer?

- kropot72

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\[\frac{1}{3}\times \frac{1}{6}=\frac{1}{18}\]

- anonymous

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

- experimentX

toss a coin ... lol

- anonymous

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

- anonymous

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

- anonymous

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 \(\frac 1 {11} .\)

- anonymous

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 \(\frac 1 6 \)

- anonymous

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

- kropot72

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

- anonymous

I think it's \(\frac {11}{36} \)

- anonymous

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

- kropot72

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

- kropot72

@FoolForMath Do you intend to reply?

- anonymous

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.

- kropot72

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.

- Zarkon

"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
\[P(E|T)=\frac{P(T|E)P(E)}{P(T)}=\frac{\frac{1}{6}\times\frac{1}{6}}{\frac{11}{36}}=\frac{1}{11}\]

- kropot72

@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.

- Zarkon

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

- kropot72

@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

- Zarkon

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

- kropot72

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.

- Zarkon

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

- anonymous

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

- Zarkon

"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?

- anonymous

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

- Zarkon

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

- kropot72

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

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