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Find the probability of AT LEAST three 6s occurring when rolling five six-sided dice all at onc

kevin2000

New member
Apr 30, 2019
1
I've been teaching myself Probability Mathematics, but I'm still struggling.

Please help me understand with an example.
Say I roll five six-sided dice all at once, with die faces numbered 1 through 6.
I want to determine the probability of AT LEAST three 6s occurring. First, is this the best way to set up the equation?

P(at least three 6s) = 1 - P(zero 6s) - P(exactly one 6) - P(exactly two 6s)

Or can the equation be better setup by subtracting AT LEAST occurrences?

I know this much...
P(zero 6s) = 5/6 * 5/6 * 5/6 * 5/6 * 5/6 = 3,125/7,776 = 0.40188
P(all 6s) = 1/6 * 1/6 * 1/6 *1/6 * 1/6 = 1/7,776
P(at least one 6) = 1 - P(zero 6s) = 4,651/7,776

To restate my whole problem, I'm unclear on the best way to calculate for "exactly 2" occurrences. Likewise for "exactly 3" or any other "exact" occurrence more than one but less than equal to the total number of dice thrown.

Thanks for your help!
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I wouldn't subtract. I would add. Rolling 5 dice, "at least three sixes" means 3 or 4 or 5. Calculate each of those separatelt then add them. On any one toss, the probability of a "six" is 1/6, the probability of anything else is 5/6.

a) Exactly 3 sixes. The probability of 3 sixes and 2 non-sixes in that order is (1/6)(1/6)(1/6)(5/6)(5/6). Writing "s" for a six, "n" for a non-six, that would be "sssnn". But there are $\frac{5!}{3!2!}= 10$ different orders in which we can write "3 s's and 2 n's and each order has the same probability so there are 10(1/6)(1/6)(1/6)(5/6)(5/6)= [FONT=Verdana,Arial,Tahoma,Calibri,Geneva,sans-serif]0.0321502[/FONT][FONT=Verdana,Arial,Tahoma,Calibri,Geneva,sans-serif] approximately.

b) Exactly 4 sixes.
The probability of 4 sixes and 1 non-six in that order is (1/6)(1/6)(1/6)(1/6)(5/6). Writing "s" for a six, "n" for a non-six, that would be "ssssn". But there are $\frac{5!}{4!1!}= 5$ different orders in which we can write "3 s's and 2 n's and each order has the same probability so there are 5(1/6)(1/6)(1/6)(1/6)(5/6)= [FONT=Verdana,Arial,Tahoma,Calibri,Geneva,sans-serif]0.0032150 [/FONT]
[FONT=Verdana,Arial,Tahoma,Calibri,Geneva,sans-serif]
approximately.

c) Exactly 5 sixes. That is, of course, (1/6)(1/6)(1/6)(1/6)(1/6)= 0.0001286, approximately.

Add those.
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