Probability (1 Viewer)

[wrxsti]

A die numbered 1 to 6 is rolled twice. The Sum of the uuppermost numbers is added to give a score....

It is known that a 5 appears on the die at least once in the two throws.. Find the probability that the Sum is greater than 9...

I got 3/11 but the answers have 3/36 ....... isnt it supposed to be 3/11 since your not taking the whole sample space now? your only taking the space with the "5's" :S

 

[Forbidden.]

For those type of questions, I draw an addition grid.
You go 1, 2, 3, 4, 5, 6 on the horizontal axis then 1, 2, 3, 4, 5, 6 on the vertical axis.
Add the scores corresponding to those numbers.
e.g 5 then 6 gives 11 then 2 and 3 gives 5 and so on ...

That's how I solve those questions. It's much easier that way.

 

[webby234]

A die numbered 1 to 6 is rolled twice. The Sum of the uuppermost numbers is added to give a score....

It is known that a 5 appears on the die at least once in the two throws.. Find the probability that the Sum is greater than 9...

I got 3/11 but the answers have 3/36 ....... isnt it supposed to be 3/11 since your not taking the whole sample space now? your only taking the space with the "5's" :S

I agree with you. Answers wrong?

 

[sangboi]

I thought i'd draw up a table.

Since its two dies being thrown these are the possible outcomes:
<table x:str="" style="border-collapse: collapse; width: 336pt;" border="0" cellpadding="0" cellspacing="0" width="448"><col style="width: 48pt;" span="7" width="64"> <tbody><tr style="height: 12.75pt;" height="17"> <td style="height: 12.75pt; width: 48pt;" align="right" height="17" width="64">+
</td> <td style="width: 48pt;" x:num="" align="right" width="64">1</td> <td style="width: 48pt;" x:num="" align="right" width="64">2</td> <td style="width: 48pt;" x:num="" align="right" width="64">3</td> <td style="width: 48pt;" x:num="" align="right" width="64">4</td> <td style="width: 48pt;" x:num="" align="right" width="64">5</td> <td style="width: 48pt;" x:num="" align="right" width="64">6</td> </tr> <tr style="height: 12.75pt;" height="17"> <td style="height: 12.75pt;" x:num="" align="right" height="17">1</td> <td class="xl22" x:num="" align="right">2</td> <td x:num="" align="right">3</td> <td x:num="" align="right">4</td> <td x:num="" align="right">5</td> <td x:num="" align="right">6</td> <td x:num="" align="right">7</td> </tr> <tr style="height: 12.75pt;" height="17"> <td style="height: 12.75pt;" x:num="" align="right" height="17">2</td> <td x:num="" align="right">3</td> <td x:num="" align="right">4</td> <td x:num="" align="right">5</td> <td x:num="" align="right">6</td> <td x:num="" align="right">7</td> <td x:num="" align="right">8</td> </tr> <tr style="height: 12.75pt;" height="17"> <td style="height: 12.75pt;" x:num="" align="right" height="17">3</td> <td x:num="" align="right">4</td> <td x:num="" align="right">5</td> <td x:num="" align="right">6</td> <td x:num="" align="right">7</td> <td x:num="" align="right">8</td> <td x:num="" align="right">9</td> </tr> <tr style="height: 12.75pt;" height="17"> <td style="height: 12.75pt;" x:num="" align="right" height="17">4</td> <td x:num="" align="right">5</td> <td x:num="" align="right">6</td> <td x:num="" align="right">7</td> <td x:num="" align="right">8</td> <td x:num="" align="right">9</td> <td x:num="" align="right">10</td> </tr> <tr style="height: 12.75pt;" height="17"> <td style="height: 12.75pt;" x:num="" align="right" height="17">5</td> <td x:num="" align="right">6</td> <td x:num="" align="right">7</td> <td x:num="" align="right">8</td> <td x:num="" align="right">9</td> <td x:num="" align="right">10</td> <td x:num="" align="right">11</td> </tr> <tr style="height: 12.75pt;" height="17"> <td style="height: 12.75pt;" x:num="" align="right" height="17">6</td> <td x:num="" align="right">7</td> <td x:num="" align="right">8</td> <td x:num="" align="right">9</td> <td x:num="" align="right">10</td> <td x:num="" align="right">11</td> <td x:num="" align="right">12</td> </tr> </tbody></table>

Since either one of the die is known to be a '5' it can have the above ^ outcomes (coloured orange/red). and since it asks for sum greater than a 9 it can be 10, 10 or 11(the red coloured ones) which is 3 out of the 36 outcomes.

Hope that helps!

** actually i noticed that the questions says 'it is known that a five appears once' that makes it impossible to have an outcome of say 1+1=2 so i do agree and think that it should be 3/11 sorry about the confusion **

 

[Ahmed A]

Yeah i got 3/36 also.
drawing the diagram which in this case seemed to be a major help in understanding the question better, but this is the first ive heard of only taking only the spaces with the fives in them...
i think regardless u still have to take all the sample space into account...

 

[webby234]

No, given that you have the knowledge that at least one die is a five, you should only take into account those possibilities.

Your answer is correct to the following question:
What is the probability that there is at least one five and the sum is greater than 9?

That is completely different to what is asked.

 

[Ahmed A]

No, given that you have the knowledge that at least one die is a five, you should only take into account those possibilities.

Your answer is correct to the following question:
What is the probability that there is at least one five and the sum is greater than 9?

That is completely different to what is asked.

come again

 

[Hikari Clover]

回复: Re: Probability

i remember this Qs......

 

[milton]

Re: 回复: Re: Probability

its called conditional probability.

you KNOW that there is at least one 5 so the sample space is reduced to 11 (equally likely) outcomes.

ignoring information will give you wrong answers