A pair of fair six-sided dice is thrown eight times. Find the probability that a score greater than #7# is scored no more than five times?

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Feb 10, 2018

Answer:

#~=0.9391#

Explanation:

Before we get into the question itself, let's talk about the method for solving it.

Let's say, for instance, that I want to account for all the possible results from flipping a fair coin three times. I can get HHH, TTT, TTH, and HHT.

The probability of H is #1/2# and the probability for T is also #1/2#.

For HHH and for TTT, that is #1/2xx1/2xx1/2=1/8# each.

For TTH and HHT, it's also #1/2xx1/2xx1/2=1/8# each, but since there are 3 ways I can get each result, it ends up being #3xx1/8=3/8# each.

When I sum up these results, I get #1/8+3/8+3/8+1/8=1# - which means I now have all the possible results of the coin flip accounted for.

Notice that if I set #H# to be #p# and therefore have #T# be #~p#, and also notice that we have a line from the Pascal's Triangle #(1,3,3,1)#, we've set up a form of:

#sum_(k=0)^(n)C_(n,k)(p)^k((~p)^(n-k))#

and so in this example, we get:

#=C_(3,0)(1/2)^0(1/2)^3+C_(3,1)(1/2)^1(1/2)^2+C_(3,2)(1/2)^2(1/2)^1+C_(3,3)(1/2)^3(1/2)^0#

#=1(1)(1/8)+3(1/2)(1/4)+3(1/4)(1/2)+1(1/8)(1)#

#=1/8+3/8+3/8+1/8=1#

Now we can do the problem.

We're given the number of rolls as 8, so #n=8#.

#p# is the sum greater than 7. To find the probability of getting a sum greater than 7, let's look at the possible rolls:

#((color(white)(0),ul1,ul2,ul3,ul4,ul5,ul6),(1|,2,3,4,5,6,7),(2|,3,4,5,6,7,8),(3|,4,5,6,7,8,9),(4|,5,6,7,8,9,10),(5|,6,7,8,9,10,11),(6|,7,8,9,10,11,12))#

Out of 36 possibilities, 15 rolls give a sum greater than 36, giving a probability of #15/36=5/12#.

With #p=5/12, ~p=7/12#

We can write out the entire sum of possibilities - from getting all 8 rolls being a sum greater than 7 all the way to getting all 8 rolls being a sum of 7 or less:

#=C_(8,0)(5/12)^8(7/12)^0+C_(8,1)(5/12)^7(7/12)^1+C_(8,2)(5/12)^6(7/12)^2+C_(8,3)(5/12)^5(7/12)^3+C_(8,4)(5/12)^4(7/12)^4+C_(8,5)(5/12)^3(7/12)^5+C_(8,6)(5/12)^2(7/12)^6+C_(8,7)(5/12)^1(7/12)^7+C_(8,8)(5/12)^0(7/12)^8=1#

but we're interested in summing up only those terms that have our greater than 7 sum happening 5 times or less:

#=C_(8,3)(5/12)^5(7/12)^3+C_(8,4)(5/12)^4(7/12)^4+C_(8,5)(5/12)^3(7/12)^5+C_(8,6)(5/12)^2(7/12)^6+C_(8,7)(5/12)^1(7/12)^7+C_(8,8)(5/12)^0(7/12)^8#

#~=0.9391#

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Feb 10, 2018

Answer:

#0.93906#

Explanation:

#"So P[ outcome > 7 ]= 15/36 = 5/12"#
#P["it occurs k times on 8 throws"] = C(8,k) (5/12)^k (7/12)^(8-k)"#
#"(binomial distribution)"#
#"with "C(n,k) = (n!)/((n-k)! k!) " (combinations)"#
#"So, "#
#P["it occurs at most 5 times on 8 throws"]#
#= 1 - P["it occurs 6, 7, or 8 times on 8 throws"]#
#= 1-C(8,6)(5/12)^6 (7/12)^2-C(8,7) (5/12)^7(7/12)-(5/12)^8#
#= 1 - (5/12)^8 (1 + 8*(7/5) + 28*(7/5)^2)#
#= 1 - (5/12)^8 (1 + 11.2 + 54.88) = 1 - (5/12)^8 (67.08)#
#= 0.93906#

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