anonymous
  • anonymous
Two die are rolled. Find the probability that the numbers are the same given that at least one is 3?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1338882073720:dw|
anonymous
  • anonymous
can you expand your question, or write it in another way..
anonymous
  • anonymous
I did not get it..

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anonymous
  • 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
  • anonymous
|dw:1338882463299:dw|
anonymous
  • anonymous
6/36 prob. of the same number
anonymous
  • anonymous
|dw:1338882580009:dw|
anonymous
  • anonymous
at least one is 3 11/36
anonymous
  • anonymous
that's all I can do, sorry..
anonymous
  • 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
  • anonymous
@experimentX : I didnot understand what you did. Would you please explain in another way?
anonymous
  • anonymous
So what would be the final answer?
kropot72
  • 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
  • anonymous
Everyone has posted a different answer. Who should I trust? :(
experimentX
  • experimentX
toss a coin ... lol
anonymous
  • anonymous
No, seriously. I'm more confused than ever before. :(
anonymous
  • 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
  • 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
  • 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
  • anonymous
I can't see any blemish in either of these approaches, however I prefer #2. May be @Zarkon can see something else :)
kropot72
  • kropot72
If a pair of dice are rolled what is the probability that one die will roll a three?
anonymous
  • anonymous
I think it's \(\frac {11}{36} \)
anonymous
  • anonymous
assuming at-least, if it is strictly one then it is \(\large\frac{10}{36}\)
kropot72
  • 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
  • kropot72
@FoolForMath Do you intend to reply?
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • Zarkon
these events are not mutually exclusive. the way you are doing it you are double counting (3,3)
kropot72
  • 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
  • Zarkon
you are describing independence ...the events are independent but they are not mutually exclusive.
anonymous
  • 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
  • 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
  • anonymous
@ Zarkon, I think it means at least one of the numbers is 3. What do u think?
Zarkon
  • 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
  • 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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