## Another roll of the…

Posted by DP on March 18, 2011 – 11:01 am

This time we have only a single {fair} die and roll it exactly 5 times. What is the probability that you will roll exactly:

a) [2] 3’s?

b) [4] 6’s?

c) [0] 1’s?

Posted by DP on March 18, 2011 – 11:01 am

This time we have only a single {fair} die and roll it exactly 5 times. What is the probability that you will roll exactly:

a) [2] 3’s?

b) [4] 6’s?

c) [0] 1’s?

March 18th, 2011 at 11:52 am

First stab at it.

a) 1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6

b) 1/6 * 1/6 * 1/6 * 1/6 * 5/6 * 5/6

c) 5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6

March 18th, 2011 at 12:23 pm

a) (1/6)^2 * (5/6)^4 * C(6,2) = 15*625/46656 ≈ 0.2009

b) (1/6)^4 * (5/6)^2 * C(6,4) = 15*25/46656 ≈ 0.008038

c) (5/6))^6 = 15625/46656 ≈ 0.3349

where C(6,2) = C(6,4) = 6!/(4! 2!) = 15

March 18th, 2011 at 12:48 pm

You may both consider looking back over the problem…

March 18th, 2011 at 12:56 pm

Ooops. 5 rolls, we did 6. (Thanks for fixing my typos, DP).

a) (1/6)^2 * (5/6)^3 * C(5,2) = 10*125/7776 ≈ 0.1608

b) (1/6)^4 * (5/6) * C(5,4) = 5*5/7776 ≈ 0.0032

c) (5/6)^5 = 3125/7776 ≈ 0.4019

March 21st, 2011 at 10:36 am

Yep. Chris has it.

This is an example of Binomial Probability.

x: # of successes

n: # of trials

p: probability of success

q: probability of failure ( = 1 – p )

b(x;n,p) = C(n,x) * p^x * q^(n-x)

where C(n,x) = P(n,x)/x!

where P(n,x) = n!/(n-x)!

so b(x;n,p) = n!/[(n-x)! x!] * p^x * q^(n-x)

For all three cases we roll the die 5 times, so n=5.

And because we are only looking for 1 specific numbre out of the 6, p=1/6 and q=5/6.

for our problems, b(x;5,1/6) = 5!/[(5-x)! x!] * (1/6)^x * (5/6)^(5-x)

a) when x=2; b(2;5,1/6) = 5!/(3! 2!) * (1/6)^2 * (5/6)^3 = 1250/7776 or 0.1607510…

b) when x=4; b(4;5,1/6) = 5!/(1! 4!) * (1/6)^4 * (5/6)^1 = 25/7776 or 0.003215…

c) when x=0; b(0;5,1/6) = 5!/(0! 5!) * (1/6)^0 * (5/6)^5

since 0!=1 and p^0=1, we get 5!/5! * 1 * (5/6)^5 = 25/7776 or 0.40187757…

I can’t figure out how to make the “roughly equal to” symbol like Chris has. Oh well, you get the point.

March 21st, 2011 at 10:41 am

cazayoux,

I had to do some brushing up on my probability & statistics on this one to figure out why to solve it this way. I would have probably tried it exactly how you did had someone else posted this problem before I read up on it.

I’m assuming you made the same mistake Chris did on rolling 6 times rather than 5. Otherwise you would have had the answer to c correct, but only by accident.

I’m not sure exactly how to put in words why we need that first term, so maybe someone else can help me out with that.

March 21st, 2011 at 3:38 pm

Hi DP. I hadn’t realised it was the binomial distribution until a while after posting. But for two 3′a I simply realised that I would get the result if I threw 33xxx,3×3xx,3xx3x,etc. That C(5,2) ways altogether. It’s only a formailty to go on to define the binomial distribution from there.

To get e.g. the “≈” symbol, I originally used charmap. But I keep a bunch of such symbols in a notepad file. The one I just used, I copy/pasted from my post above.

March 26th, 2011 at 8:59 pm

i think the ans is-[4] 6’s

but i’m not fully sure about it……..

becoz out of five chances the probablity of getting a fair no is 4 and out of 6 different no’s……..

March 28th, 2011 at 8:14 am

Thanks for the help Chris.

rao…I’m not sure how to respond to you. I don’t see an answer, so I can’t tell you you are wrong.

I think you may have thought it was a multiple choice question. It is a multiple part question, which you must answer each part (a, b, & c) with a probability.

Your last sentence fragment isn’t very clear to me. What are you refering to when you say “fair no”?