text: a bunch of 12 flowers has three types of flowers, tulips, daffodils and roses.
  
  4 of the flowers are tulips.
  3 of the flowers are daffodils.
  5 of the flowers are roses
  
  James taes a flower at random from the bunch.
  He does not replace it.
  he then takes a second flower at random from the bunch.
  
  work out the probability that ther two flowers will NOT be of the same type.
  
  any help please?? and how can i show my working?
  A tree diagram would be best for this. Remember that the 2n probability changes because he does not replace it
  Sunderland AFC
  how? lol
  
  YOu would write the answer like: (I think)
  
  (12C1)(11C1)
  ____________
  
  (12C2)
  
  Im not completely sure about this, but I do know that you need to use the "C" function, since this is a combination equation.
  Its quite easy, you just take one of the flowertypes, tullips for instance, the chance the first one is a tullip is 4/12 the chance the next one isn't a tullip is 8/11, then you take another flowertype daffodils the chance the first one is a daffodil is 3/12 the chance the second one isn't is 9/11, then you take the last type of flower which is roses the chance the first you pick is a rose is 5/12 the chance the second isn't is 7/11 so you get 4/12*8/11+3/12*9/11+5/12*7/11 there are more official ways to calculate it but in cases like this i just calculate it this way
  You would have three branches of your tree with the labels and the probability (i.e 4/12, 3/12 and 5/12) accordingly. Then have three branches coming out of ach of these three all labelelled but this time the probability will have changed. eg if he picked out a tulip then you would follow the tulip branch and the new prob for tulips on the next set of branches would be 3/11. however the others will remain the same on that set of branches. This applies for all other branches and then you would multiply the two fractions at the end.
  This would be much easier if i could draw it for you.
  You would then add together all the fractions were the two picked were different and ths woul;d give you the final probability.
  Sunderland AFC
  Instruction by example.
explanation: The text extract contains a math question and multiple responses providing answers and explanations to the question. The question asks for the probability that two flowers randomly picked from a bunch of 12 flowers (with 4 tulips, 3 daffodils, and 5 roses) will not be of the same type. The responses include different methods to solve the problem, such as using combinations and tree diagrams.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: a bunch of 12 flowers has three types of flowers, tulips, daffodils and roses.
  
  4 of the flowers are tulips.
  3 of the flowers are daffodils.
  5 of the flowers are roses
  
  James taes a flower at random from the bunch.
  He does not replace it.
  he then takes a second flower at random from the bunch.
  
  work out the probability that ther two flowers will NOT be of the same type.
  
  any help please?? and how can i show my working?
  A tree diagram would be best for this. Remember that the 2n probability changes because he does not replace it
  Sunderland AFC
  how? lol
  
  YOu would write the answer like: (I think)
  
  (12C1)(11C1)
  ____________
  
  (12C2)
  
  Im not completely sure about this, but I do know that you need to use the "C" function, since this is a combination equation.
  Its quite easy, you just take one of the flowertypes, tullips for instance, the chance the first one is a tullip is 4/12 the chance the next one isn't a tullip is 8/11, then you take another flowertype daffodils the chance the first one is a daffodil is 3/12 the chance the second one isn't is 9/11, then you take the last type of flower which is roses the chance the first you pick is a rose is 5/12 the chance the second isn't is 7/11 so you get 4/12*8/11+3/12*9/11+5/12*7/11 there are more official ways to calculate it but in cases like this i just calculate it this way
  You would have three branches of your tree with the labels and the probability (i.e 4/12, 3/12 and 5/12) accordingly. Then have three branches coming out of ach of these three all labelelled but this time the probability will have changed. eg if he picked out a tulip then you would follow the tulip branch and the new prob for tulips on the next set of branches would be 3/11. however the others will remain the same on that set of branches. This applies for all other branches and then you would multiply the two fractions at the end.
  This would be much easier if i could draw it for you.
  You would then add together all the fractions were the two picked were different and ths woul;d give you the final probability.
  Sunderland AFC
  Instruction by example.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract contains a math question and multiple responses providing answers and explanations to the question. The question asks for the probability that two flowers randomly picked from a bunch of 12 flowers (with 4 tulips, 3 daffodils, and 5 roses) will not be of the same type. The responses include different methods to solve the problem, such as using combinations and tree diagrams.
  
  Label: question-and-answer