# Binomial probability.

• Jul 16th 2011, 08:16 PM
university
Binomial probability.
a survey is carried out on a computer network.the probability that a log on to the network is succesful is 0.97.find the probabilty that exactly six out of eight users that attempt to log on will do so successfully.
x is bin(8, 0.97)
P[6] = C(8,6) * 0.97^6 * 0.03^2 = 0.0209908945242108 is this correct or am i doing this completely wrong???
• Jul 16th 2011, 08:31 PM
pickslides
Re: Probabilty help!!!!!!!!!!!!?
Quote:

Originally Posted by university
P[6] = C(8,6) * 0.97^6 * 0.03^2

Sounds good to me.
• Jul 17th 2011, 12:22 AM
mr fantastic
Re: Binomial probability.
Quote:

Originally Posted by university
a survey is carried out on a computer network.the probability that a log on to the network is succesful is 0.97.find the probabilty that exactly six out of eight users that attempt to log on will do so successfully.
x is bin(8, 0.97)
P[6] = C(8,6) * 0.97^6 * 0.03^2 = 0.0209908945242108 is this correct or am i doing this completely wrong???

The number of significant figures given is - sorry - absurd. Most questions state the accuracy required - check the question and round your answer accordingly.
• Jul 17th 2011, 08:16 PM
CaptainBlack
Re: Binomial probability.
Quote:

Originally Posted by mr fantastic
The number of significant figures given is - sorry - absurd. Most questions state the accuracy required - check the question and round your answer accordingly.

No need to check the question the clue is in the precision of the log-on probability. This is given to two significant figures, no naively we might expect that we should give the answer two significant figures or 0.021.

But wait I hear you cry, a probability of 0.97 indicates a range of possible values of 0.965 to 0.975 (give or take an end point). That gives a range for the answer of 0.0277 to 0.0150, so we have at best only one significant figure, so the answer should be truncated to 0.02.

CB
• Jul 17th 2011, 08:23 PM
mr fantastic
Re: Binomial probability.
Quote:

Originally Posted by CaptainBlack
No need to check the question the clue is in the precision of the log-on probability. This is given to two significant figures, no naively we might expect that we should give the answer two significant figures or 0.021.

But wait I hear you cry, a probability of 0.97 indicates a range of possible values of 0.965 to 0.975 (give or take an end point). That gives a range for the answer of 0.0277 to 0.0150, so we have at best only one significant figure, so the answer should be truncated to 0.02.

CB

Aha .... but maybe the log-on probability is - unbelievably, perhaps - an exact probability ....
• Jul 18th 2011, 01:40 AM
CaptainBlack
Re: Binomial probability.
Quote:

Originally Posted by mr fantastic
Aha .... but maybe the log-on probability is - unbelievably, perhaps - an exact probability ....

I'm sure that would violate some rule about how it should be written, but which I don't know!? :(

CB