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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1338882073720:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you expand your question, or write it in another way..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1338882463299:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.06/36 prob. of the same number

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1338882580009:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0at least one is 3 11/36

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's all I can do, sorry..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Are 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
 4 years ago
Best ResponseYou've already chosen the best response.0@experimentX : I didnot understand what you did. Would you please explain in another way?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So what would be the final answer?

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0Everyone has posted a different answer. Who should I trust? :(

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0toss a coin ... lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, seriously. I'm more confused than ever before. :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sample set in a role of two dice: $$ \begin{array}{cccC} &1&2&3&4&5&6 \\ \hline 1 &(1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\ \hline 2 &(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\ \hline 3 &(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\ \hline 4 &(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\ \hline 5 &(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\ \hline 6 &(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\\ \hline \end{array}$$

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Observe 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
 4 years ago
Best ResponseYou've already chosen the best response.0You 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
 4 years ago
Best ResponseYou've already chosen the best response.0I can't see any blemish in either of these approaches, however I prefer #2. May be @Zarkon can see something else :)

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.0If a pair of dice are rolled what is the probability that one die will roll a three?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think it's \(\frac {11}{36} \)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0assuming atleast, if it is strictly one then it is \(\large\frac{10}{36}\)

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0@FoolForMath Do you intend to reply?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0They 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
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.2"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 \[P(3_1 \text{ or } 3_2)=1P(\left(3_1 \text{ or } 3_2\right)')=1P(3'_1 \text{ and } 3'_2)\] \[=1\frac{5}{6}\times\frac{5}{6}=1\frac{25}{36}=\frac{11}{36}\] the answer to the OP is \(\displaystyle\frac{1}{11}\) E=equal , T=at least one 3 \[P(ET)=\frac{P(TE)P(E)}{P(T)}=\frac{\frac{1}{6}\times\frac{1}{6}}{\frac{11}{36}}=\frac{1}{11}\]

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.0@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
 4 years ago
Best ResponseYou've already chosen the best response.2the condition, from the OP, is 'given that at least one is 3' not 'only one die of the pair will have a three'

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.0@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
 4 years ago
Best ResponseYou've already chosen the best response.2these events are not mutually exclusive. the way you are doing it you are double counting (3,3)

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.0The number that comes up on die #1 has no bearing on the number that comes up on die #2. The subevents '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
 4 years ago
Best ResponseYou've already chosen the best response.2you are describing independence ...the events are independent but they are not mutually exclusive.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.2"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
 4 years ago
Best ResponseYou've already chosen the best response.0@ Zarkon, I think it means at least one of the numbers is 3. What do u think?

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2"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
 4 years ago
Best ResponseYou've already chosen the best response.0@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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