Probability Question?

Suppose you roll a six-sided die 10 times to create three sets of numbers. The first set contains rolls 1 - 3 (R1 - R3) and the second set contains rolls 4 - 6 (R4 - R6), and the third set contains rolls 7 - 9 (R7 - R9). Each set then contains a shared fourth number, the 10th roll of the die. Each set then is given by:

Set 1 = {R1, R2, R3, R10}

Set 2 = {R4, R5, R6, R10}

Set 3 = {R7, R8, R9, R10}

What is the probability that at least one of the sets does not contain the number six?

Thanks so much for any help on this :)

2 Answers

  • atsuo
    Lv 6
    1 month ago

    Given 3 sets : 

    Set1 = {R1, R2, R3, R10} 

    Set2 = {R4, R5, R6, R10} 

    Set3 = {R7, R8, R9, R10} 


    Think 3 subsets Sx,Sy,Sz such that 

    Sx = {R1, R2, R3} 

    Sy = {R4, R5, R6} 

    Sz = {R7, R8, R9} 


    So given condition can be modified as 

    "At least one of given 3 sets does not contain 6" 

    = "At least one of Sx,Sy,Sz does not contain 6" and "R10 is not 6"  

    = "Each of Sx,Sy,Sz contains 6" does not occur and "R10 is not 6" ---(#1) 


    We know "Sx contains 6" = "Each of R1,R2,R3 is not 6" does not occur , 

    so its probability is  

    1 - (5/6) * (5/6) * (5/6) 

    = 1 - 125/216 

    = 91/216 


    Similarly , the probabilities of "Sy contains 6" and "Sz contains 6" are 

    91/216 and 91/216 . 


    So the probability of "Each of Sx,Sy,Sz contains 6" is 

    (91/216)^3 = 753571/10077696 . 


    So the probability that "Each of Sx,Sy,Sz contains 6" does not occur is 

    1 - 753571/10077696 = 9324125/10077696 . 


    By (#1) , the answer is 

    (9324125/10077696) * (5/6) = 46620625/60466176 .

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  • 1 month ago

    There might eb a formula, but I always like to think through a problem like this logically, rather than memorize however many formula there can be  Probability can get super specific.  

    Since the tenth roll is special let'slook at the first 9.  How many possible results are there?  so for instance something like {5,6,6,3,5,5,3,4,5} could be a result of the first 9 rolls.  How many are there total? 10,077,696.  If youa re unaware you get this by doing 6^9 since the dice has 6 faces and you are rolling 9 times.  or you could start at 1.  1 roll has 6 possibilities, 2 rolls has 36 and so on, you just keep multiplying by 6.  it may be easier if you use a coin to really count how many results there could be with repeated rolls/ flips, but back to the question.

    Now of those 10,077,696, how many have the first three rolls not have a 6?  How many have no 6s at all?  That's what we're going to find out.  

    There are 7 stuations we can look at if we want at least one set to not have a 6

    Set 1 is the only one without a 6

    Set 2 is the only one without a 6

    Set 3 is the only one without a 6

    Only Set 1 has a 6

    Only set 2 has a 6

    Only set 3 has a 6

    None of them have a 6

    Let's work from the bottom up since I think that's the easiest.

    Calculating the probability of none of the 10 rolls having a 6 is pretty straightforward.What's the probability of not getting a 6 with the first roll?  5/6.  how about in two rolls?  5/6 * 5/6.  You just keep multiplying by 5/6 for every roll you don't want 6 in.  so in the end it's (5/6)^10.  This is 9765625/60466176.  Or in other words, rolling a dice 10 times has 60466176 possibilities.  And of those there are 9765625 that do not have a single 6 in them.  

    Now let's work on if only set 3 has a 6.  We can do this just like the other one up through roll 6, but when we get to roll 7 we have to change it up.  so one roll is 5/6, two rolls is 5/6 * 5/6 all the way to 6 rolls is (5/6)^6 = 15625/46656, or again, after 6  rolls there are 46656 total possibilities and of them 15625 have no 6s  Now with rolls 7,8 ad 9 we have to make sure there is at leat one 6.  The trick is that there could be more than  too, so I am going to find the probability of each combination.  Those being just roll 7 having one, just roll 8, just roll 9, both rolls 7 and 8, just rolls 7 and 9, just rolls 8 and 9 and all three rolls.  Since there are these  different possibilities, instead of multiplying each individual roll let's just find the probability of these three rolls having at least one 6 and use that as one term.

    So what is the probability of there being at least one 6 in three rolls?  I find the probability of each combination.

    All three is the most straight forward, again it's just multiplying the probability for each roll.  So what is the probability of getting a 6 on one roll?  1/6.  Two rolls is 1/6 * 1/6 and then three rolls is 1/6 * 1/6 * 1/6.  Then we use similar calculations to find the probabilities for the other options.

    Just roll 7 = 1/6 * 5/6 * 5/6 = 25/216

    Just roll 8 = Smae as 4

    Just roll 9 = Same as 4

    Roll 7 and 8 = 1/6 * 1/6 * 5/6 = 5/216

    Roll 7 and 9 = 5/216

    Roll 8 and 9 = 5/216

    All three rolls = 1/216

    Now you can add these together to get the probabilty of getting at least one 6 in three rolls, so 91/216.

    NOW going back to the (5/6)^6 we can multiply this by 91/216, and this gets us the probability that after 9 rolls only rolls 7, 8 and 9 have a 6, and it can be any three of them, including all three.  Of course there's one more roll, but since roll 10 is accounted for in all the sets we NEVER want a 6 in roll 10, so then just multiply the other two parts by 5/6, so in total that leaves us with (5/6)^6(91/216)(5/6) = 7109375/60466176 Or in other words, after rolling 10 times there are again 60466176 total possibilities, and of them 7109375 have 6s only appear in roll 7, 8 or 9.

    So now we have two parts, we have the probability of there being no 6s at all, and now only 6s in set 3.  As you can see writing out the math is not fun, but the thought process is hopefully intuitive.  This 7109375/60466176 is the same for a 6 in only set 2 and 3 as well, so now we know that the probability of getting no 6s or a 6 in just one set is 31093750/60466176

    Can you manage the probabilities of just one set not having a 6?  i could walk you through finding that.   But in the end you just add the probabilities up.  If you don't understand why they are getting added up let me know as well.  

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