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Finding probability that two chosen numbers have a multiple of nine.

Urn Probability (need multiple ways)Probability a product of $n$ randomly chosen numbers from 1-9 is divisible by 10.Two numbers are chosen at random over the interval $ [0,1]$What is the probability that a randomly chosen positive three-digit integer is a multiple of $7$?Probability - What is the probability that the two-digit number is a multiple of 3Probability of choosing same multipleProbability of Hearts given Hearts Was ChosenWhat is the probability that three randomly chosen computers were not tested?Probability that a randomly chosen element from set $A = {1, 2, 3, ldots, 1000}$ is a multiple of $2$ OR $3$.A lottery will be held. From 1000 numbers, one will be randomly chosen as the winner.

0

$begingroup$

If two numbers will be randomly chosen without replacement from ${3, 4, 5, 6}$, what is the probability that their product will be a multiple of 9?

What I did:
To be a multiple of 9, a number needs to have 2 threes. The only possible choice in the given set is choosing 3 and 6. Choosing either 3 or 6 has a probability of $frac{1}{4}$, and choosing the other number that's left has a probability of $frac{1}{3}$ which gives $frac{1}{4}cdotfrac{1}{3}=frac{1}{12}$, but this is wrong. What did I do wrong?

probability

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asked Mar 13 at 23:31

Max0815Max0815

77518

$endgroup$

add a comment | 

0

$begingroup$

If two numbers will be randomly chosen without replacement from ${3, 4, 5, 6}$, what is the probability that their product will be a multiple of 9?

What I did:
To be a multiple of 9, a number needs to have 2 threes. The only possible choice in the given set is choosing 3 and 6. Choosing either 3 or 6 has a probability of $frac{1}{4}$, and choosing the other number that's left has a probability of $frac{1}{3}$ which gives $frac{1}{4}cdotfrac{1}{3}=frac{1}{12}$, but this is wrong. What did I do wrong?

probability

share|cite|improve this question

asked Mar 13 at 23:31

Max0815Max0815

77518

$endgroup$

add a comment | 

$begingroup$

If two numbers will be randomly chosen without replacement from ${3, 4, 5, 6}$, what is the probability that their product will be a multiple of 9?

What I did:
To be a multiple of 9, a number needs to have 2 threes. The only possible choice in the given set is choosing 3 and 6. Choosing either 3 or 6 has a probability of $frac{1}{4}$, and choosing the other number that's left has a probability of $frac{1}{3}$ which gives $frac{1}{4}cdotfrac{1}{3}=frac{1}{12}$, but this is wrong. What did I do wrong?

probability

share|cite|improve this question

asked Mar 13 at 23:31

Max0815Max0815

77518

$endgroup$

share|cite|improve this question

asked Mar 13 at 23:31

Max0815Max0815

77518

share|cite|improve this question

asked Mar 13 at 23:31

Max0815Max0815

77518

asked Mar 13 at 23:31

Max0815Max0815

77518

asked Mar 13 at 23:31

Max0815Max0815

77518

2 Answers
2

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oldest

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2

$begingroup$

You counted the case choosing $3$ and then $6$ and choosing $6$ and then $3$ as $2$ different cases. So the answer should be $frac{1}{6}$. The order does not matter as long as you choose $3$ and $6$.

Alternatively, number of elements in your sample space is $4choose 2$, which is $6$, with only $1$ event of interest.

share|cite|improve this answer

edited Mar 13 at 23:41

answered Mar 13 at 23:38

ShaileshShailesh

3,98092134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh! Oof I fail at dis lol xD! Thanks! :P
  $endgroup$
  – Max0815
  Mar 13 at 23:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No worries, double counting is a common mistake
  $endgroup$
  – Shailesh
  Mar 13 at 23:41
  

  
  
  
  

add a comment | 

1

$begingroup$

The first choice has to be a $3$ or a $6$, with probability $frac{1}{2}$. The other number is chosen with a probability $frac{1}{3}$, net probability is $frac{1}{6}$.

share|cite|improve this answer

answered Mar 13 at 23:40

herb steinbergherb steinberg

3,0942311

$endgroup$

add a comment | 

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2

$begingroup$

You counted the case choosing $3$ and then $6$ and choosing $6$ and then $3$ as $2$ different cases. So the answer should be $frac{1}{6}$. The order does not matter as long as you choose $3$ and $6$.

Alternatively, number of elements in your sample space is $4choose 2$, which is $6$, with only $1$ event of interest.

share|cite|improve this answer

edited Mar 13 at 23:41

answered Mar 13 at 23:38

ShaileshShailesh

3,98092134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh! Oof I fail at dis lol xD! Thanks! :P
  $endgroup$
  – Max0815
  Mar 13 at 23:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No worries, double counting is a common mistake
  $endgroup$
  – Shailesh
  Mar 13 at 23:41
  

  
  
  
  

add a comment | 

2

$begingroup$

You counted the case choosing $3$ and then $6$ and choosing $6$ and then $3$ as $2$ different cases. So the answer should be $frac{1}{6}$. The order does not matter as long as you choose $3$ and $6$.

Alternatively, number of elements in your sample space is $4choose 2$, which is $6$, with only $1$ event of interest.

share|cite|improve this answer

edited Mar 13 at 23:41

answered Mar 13 at 23:38

ShaileshShailesh

3,98092134

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh! Oof I fail at dis lol xD! Thanks! :P
  $endgroup$
  – Max0815
  Mar 13 at 23:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No worries, double counting is a common mistake
  $endgroup$
  – Shailesh
  Mar 13 at 23:41
  

  
  
  
  

add a comment | 

$begingroup$

You counted the case choosing $3$ and then $6$ and choosing $6$ and then $3$ as $2$ different cases. So the answer should be $frac{1}{6}$. The order does not matter as long as you choose $3$ and $6$.

Alternatively, number of elements in your sample space is $4choose 2$, which is $6$, with only $1$ event of interest.

share|cite|improve this answer

edited Mar 13 at 23:41

answered Mar 13 at 23:38

ShaileshShailesh

3,98092134

$endgroup$

share|cite|improve this answer

edited Mar 13 at 23:41

answered Mar 13 at 23:38

ShaileshShailesh

3,98092134

answered Mar 13 at 23:38

ShaileshShailesh

3,98092134

answered Mar 13 at 23:38

ShaileshShailesh

3,98092134

1

$begingroup$

The first choice has to be a $3$ or a $6$, with probability $frac{1}{2}$. The other number is chosen with a probability $frac{1}{3}$, net probability is $frac{1}{6}$.

share|cite|improve this answer

answered Mar 13 at 23:40

herb steinbergherb steinberg

3,0942311

$endgroup$

add a comment | 

1

$begingroup$

The first choice has to be a $3$ or a $6$, with probability $frac{1}{2}$. The other number is chosen with a probability $frac{1}{3}$, net probability is $frac{1}{6}$.

share|cite|improve this answer

answered Mar 13 at 23:40

herb steinbergherb steinberg

3,0942311

$endgroup$

add a comment | 

$begingroup$

The first choice has to be a $3$ or a $6$, with probability $frac{1}{2}$. The other number is chosen with a probability $frac{1}{3}$, net probability is $frac{1}{6}$.

share|cite|improve this answer

answered Mar 13 at 23:40

herb steinbergherb steinberg

3,0942311

$endgroup$

share|cite|improve this answer

answered Mar 13 at 23:40

herb steinbergherb steinberg

3,0942311

answered Mar 13 at 23:40

herb steinbergherb steinberg

3,0942311

answered Mar 13 at 23:40

herb steinbergherb steinberg

3,0942311