Probability

[istar]

A bus company has records showing that its buses arrive on time 95% of the time. Suppose the company operates 65 bus trips each day. The CEO has asked for the probability that 60 or more of these buses arrive on time. Use the binomial distribution to determine the probability that 60 or more buses arrive on time.

n=65, x= 60 p=0.95, q=0.05
ncr*(p^r)*(q)^n-r = 65C60*(0.95)^60*(0.05)^65-60

I wonder if I am on the right way pls ........

 

[Steven G]

It depends on what you think you are finding the probability of to get the answer you got.

 

[istar]

Looking the probability that 60 or more buses arrive on time using binomial or by using normal approximation . So , I though by constructing with the binomial formula is the way...

 

[istar]

n=65, X= 60 p=0.95, q=0.05
p(X) = ncr*(p)^r*(q)^n-r = 65C60*(0.95)^60*(0.05)^65-60
ncr= (65!) / (65-60)!(60)! = (65*64*63*62*61*60!) / (4)!(60)!
= (65*64*63*62*61*60!) / (4*3*2*1)(60)! = (991,186,560) / 24 = 41,299,440
(p)^r*(q)^n-r = *(0.95)^60*(0.05)^65-60
= (0.046069798)*(0.0000003125)= 1.439 *10^-8
p(X) = 41,299,440 * 1.439 *10^-8 = 0.594580035 OR 59%
Thus, the probability that 60 or more buses arrive on time is 59%.

 

[Dr.Peterson]

n=65, X= 60 p=0.95, q=0.05
p(X) = ncr*(p)^r*(q)^n-r = 65C60*(0.95)^60*(0.05)^65-60
ncr= (65!) / (65-60)!(60)! = (65*64*63*62*61*60!) / (4)!(60)!
= (65*64*63*62*61*60!) / (4*3*2*1)(60)! = (991,186,560) / 24 = 41,299,440
(p)^r*(q)^n-r = *(0.95)^60*(0.05)^65-60
= (0.046069798)*(0.0000003125)= 1.439 *10^-8
p(X) = 41,299,440 * 1.439 *10^-8 = 0.594580035 OR 59%
Thus, the probability that 60 or more buses arrive on time is 59%.

I get 1/5 of your value for [MATH]_{65}C_{60}[/MATH], using a calculator's nCr button.

For p^r*q^(n-r) [note correct use of parentheses], I get what you got.

For the product I get 0.1189, which is 1/5 of your result.

But the important thing is that the formula you used gives the probability that EXACTLY 60 arrive on time. You have more work to do to answer the question: find the probability that 60, or 61, or 62, or 63, or 64, or 65 arrive on time. Or, use the normal approximation that you mentioned.