A fair six-sided die is tossed 10 times. Find the probability that the first tos

Question

A fair six-sided die is tossed 10 times. Find the probability that the first toss is a 6 given there are exactly three 6's.
If you provide a solution please show your work so I can know how to answers questions like this in the future.Thank you! A fair six-sided die is tossed 10 times. Find the probability that the first toss is a 6 given there are exactly three 6's.
If you provide a solution please show your work so I can know how to answers questions like this in the future.Thank you!
If you provide a solution please show your work so I can know how to answers questions like this in the future.Thank you!

Explanation / Answer

Let

F6 = first is a six

3S = 3 sixes

Thus,

P(3S n F6) = (1/6)*P(2 6s in 9 tries)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    9      
p = the probability of a success =    0.166666667      
x = the number of successes =    2      
          
Thus, the probability is          
          
P (    2   ) =    0.279081647

Thus,

P(3S n F6) = (1/6)*0.279081647 = 0.046513608

**************

Now, the probability of 3 6s in 10 tosses:

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    10      
p = the probability of a success =    0.166666667      
x = the number of successes =    3      
          
Thus, the probability is          
          
P (    3   ) =    0.15504536

Thus,

P(F6|3S) = 0.046513608/0.15504536 = 0.3 [ANSWER]

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