Melody

It is incorrect.   

@@ End of Day Wrap       Sat 22/11/14       Sydney,   Australia     Time 12:35 am  (Really Sunday)         ♬

Hi everyone,

Today was probably the busiest Saturday that we have ever had.  The forum is growing exponentially.

Anyway there were lots of great answers provided by Alan, CPhill, Geno3141, BabaLlama, SelenaPubblT, Mathematician, Heureka, and Happy7.  Thank you, you are all great.    

There were many questions that did not get answered properly today - many but not of them were probablility.   I collected the address and put them in this thread.  ( I appologise if there is any doubling up)

Sun 23/11/14

Finding percent of a number (or money)

I suggest you look up a table OR here is a site that you could make good use of

Should there be any brackets here?

What is the sum of the first 25 terms of the sequence 3, 17, 31, 45, ..

d=14 a=3 n=25

Sn=n/2(2a+(n-1)d) = 25/2(6+24*14) 

$${\frac{{\mathtt{25}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\left({\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{14}}\right) = {\mathtt{4\,275}}$$

The random variable X has a binomial distribution with n = 10 and p = 0.01. Determine the following probabilities.

            a. P(X = 5)                  c. P(X ≥ 9)

            b. P(X ≤ 2)                  d.P(3 ≤ X < 5)

a)  10C5*0.01^5*0.99^5

b) P(0 or 1 or 2 successes)=  0.99^10 + 10C1*0.01*0.01^9 + 10C2*0.01^2*0.99^8

c) P(9 or 10 sucesses) = 10C9*0.01^9*0.99 + 0.99^10

d) P(3 or 4 sucesses) = 10C3*0.01^3*0.99^7 + 10C4*0.01^4*0.99^6

I think these are right but it is late - i am tired and probability is not my strong suite.

The phone lines to an airline reservation system are occupied 40% of the time. Assume that the events that the lines are occupied on successive calls are independent. Assume that 10 calls are placed to the airline.

a.  What is the probability that for exactly three calls the lines are occupied?

10C3* 0.4^3*0.6^7

$${\left({\frac{{\mathtt{10}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.4}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{0.6}}}^{{\mathtt{7}}} = {\mathtt{0.214\: \!990\: \!848}}$$

b. What is the probability that for at least one call the lines are not occupied?

1-P(all are cccupied)=$${\mathtt{1}}{\mathtt{\,-\,}}{{\mathtt{0.4}}}^{{\mathtt{10}}} = {\mathtt{0.999\: \!895\: \!142\: \!4}}$$

The probability that a person carries the gene is 0.1. What is the probability that 4 will have to be tested before 2 with the gene are detected?

I think this really means that there is one sucess in the first 3 and then the forth one is a success.

P(1 succes in first 3)=3C1*(0.1)^1(0.9)^2

$${\mathtt{3}}{\mathtt{\,\times\,}}{\left({\mathtt{0.1}}\right)}^{{\mathtt{1}}}{\mathtt{\,\times\,}}{\left({\mathtt{0.9}}\right)}^{{\mathtt{2}}} = {\frac{{\mathtt{243}}}{{\mathtt{1\,000}}}} = {\mathtt{0.243}}$$

P(1 succes in first 3 FOLOWED by a success) = 0.243*0.1 = 0.0243

Once again - I think that is correct  

a) P(FFFS)=0.2^3*0.8 

$${{\mathtt{0.2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{0.8}} = {\frac{{\mathtt{4}}}{{\mathtt{625}}}} = {\mathtt{0.006\: \!4}}$$

b) 1-P(FFFF) = 1-0.2^4 

$${\mathtt{1}}{\mathtt{\,-\,}}{{\mathtt{0.2}}}^{{\mathtt{4}}} = {\frac{{\mathtt{624}}}{{\mathtt{625}}}} = {\mathtt{0.998\: \!4}}$$

c) P(FFF)=0.2^3

$${{\mathtt{0.2}}}^{{\mathtt{3}}} = {\frac{{\mathtt{1}}}{{\mathtt{125}}}} = {\mathtt{0.008}}$$

That is what I think anyway.  :))

Perhaps someone would like to check my answers?