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emcrazy14
Two die are rolled. Find the probability that the numbers are the same given that at least one is 3?
can you expand your question, or write it in another way..
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.
6/36 prob. of the same number
that's all I can do, sorry..
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?
@experimentX : I didnot understand what you did. Would you please explain in another way?
So what would be the final answer?
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\[\frac{1}{3}\times \frac{1}{6}=\frac{1}{18}\]
Everyone has posted a different answer. Who should I trust? :(
toss a coin ... lol
No, seriously. I'm more confused than ever before. :(
Sample set in a role of two dice: $$ \begin{array}{c|c|c|C|} &1&2&3&4&5&6 \\ \hline 1 &(1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\ \hline 2 &(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\ \hline 3 &(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\ \hline 4 &(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\ \hline 5 &(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\ \hline 6 &(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\\ \hline \end{array}$$
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 \(\frac 1 {11} .\)
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 \(\frac 1 6 \)
I can't see any blemish in either of these approaches, however I prefer #2. May be @Zarkon can see something else :)
If a pair of dice are rolled what is the probability that one die will roll a three?
I think it's \(\frac {11}{36} \)
assuming at-least, if it is strictly one then it is \(\large\frac{10}{36}\)
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
@FoolForMath Do you intend to reply?
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.
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.
"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 \[P(E|T)=\frac{P(T|E)P(E)}{P(T)}=\frac{\frac{1}{6}\times\frac{1}{6}}{\frac{11}{36}}=\frac{1}{11}\]
@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.
the condition, from the OP, is 'given that at least one is 3' not 'only one die of the pair will have a three'
@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
these events are not mutually exclusive. the way you are doing it you are double counting (3,3)
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.
you are describing independence ...the events are independent but they are not mutually exclusive.
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
"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?
@ Zarkon, I think it means at least one of the numbers is 3. What do u think?
"I think it means at least one of the numbers is 3." yes it doesn't mean "the probability of getting 3 and above"
@Zarkon Thank you very much for your patience and explanations. I accept that the Addition Theorem does not apply in this problem :)