What is the Probability of Throwing Two Dice and Getting Specific Numbers?

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The discussion focuses on calculating the probability of specific outcomes when throwing two dice, given that one die shows a 6. For part (a), the correct probability of the other die showing a 2 is 2/11, contrary to the initial calculation of 1/36. For part (b), the probability that the total score is greater than 8 is determined to be 7/11. The conversation highlights confusion regarding the wording of the question, which affects the interpretation of the probabilities. Clarifying the conditions under which the probabilities are calculated is crucial for accurate results.
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Homework Statement



A pair of dice are thrown. If a die show number 6 , what is the probability of (a)another die show number 2? (b) total score for both number greater than 8 ?

for part (a) , my working is 1/6 x 1/6 = 1/36

but actual ans is 2/11 for part (b), the ans is 7/11


Homework Equations





The Attempt at a Solution

 
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desmond iking said:

Homework Statement



A pair of dice are thrown. If a die show number 6 , what is the probability of (a)another die show number 2? (b) total score for both number greater than 8 ?

for part (a) , my working is 1/6 x 1/6 = 1/36

but actual ans is 2/11 for part (b), the ans is 7/11


Homework Equations





The Attempt at a Solution


Write down all the outcome pairs where at least one die shows a 6. How many such pairs do you have? How many of those pairs also include a 2?
 
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(1,6) , (2,6) , (3,6) , (4,6) , (5,6) , (6,6)

one contain (2,6)
 
What about (6,1),(6,2) etc.?
 
LCKurtz said:
What about (6,1),(6,2) etc.?

hi, i got the solution now. but i just can't understand why i can't use the probability tree to get the ans as the photo below?
 

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You drew a tree, but you haven't shown getting an answer to the question using it. What answer do you think the tree is leading you to?
 
desmond iking said:
A pair of dice are thrown. If a die show number 6 , what is the probability of (a)another die show number 2?
Is that the exact wording? If so, it's a terrible question. As LCKurtz presumed, it almost certainly intended "if at least one die shows a 6", but that's not what it says.
It all hinges on the basis for selecting the die showing a six. The wording suggests that the observer picked one die at random and saw that it was a six. That will not give 2/11 as the answer.
 

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