Central Station: prob. at least 6 trains are on time....

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65 % of all scheduled trains passing through Central Station arrive on time. If 8 trains go through the train every hour, find:

i. The probability that at least 6 trains are on time
ii. The probability that at least 6 trains are on time for 9 of the next 10 hours.

I don't want answers
just how to do it...thanks!
 
Re: Central Station

americo74 said:
65 % of all scheduled trains passing through Central Station arrive on time. If 8 trains go through the train every hour, find:

i. The probability that at least 6 trains are on time

\(\displaystyle \L\\\sum_{k=6}^{8}\begin{pmatrix}8\\k\end{pmatrix}p^{k}q^{8-k}\)

p=0.65 and q=0.35


ii. The probability that at least 6 trains are on time for 9 of the next 10 hours.

I don't want answers
just how to do it...thanks!
 
Re: Central Station

Hello, americo74!

Let me baby-talk through Galactus' explanation . . .


65 % of all scheduled trains passing through Central Station arrive on time.
If 8 trains go through the train every hour, find:

a) The probability that at least 6 trains are on time during the next hour

b) The probability that at least 6 trains are on time for 9 of the next 10 hours.

(a) We want the probability that: 6 are on time \displaystyle \text{6 are on time } or  7 are on time \displaystyle \text{ 7 are on time } or  8 are on time.\displaystyle \text{ 8 are on time.}

Find the three probabilities and add them.

    P(6 on time)=(86)(0.65)6(0.35)2\displaystyle \;\;P(\text{6 on time}) \:=\:\begin{pmatrix}8\\6\end{pmatrix}(0.65)^6(0.35)^2

    P(7 on time)=(87)(0.65)7(0.35)\displaystyle \;\;P(\text{7 on time}) \:=\:\begin{pmatrix}8\\7\end{pmatrix}(0.65)^7(0.35)

    P(8 on time)=(88)(0.65)8\displaystyle \;\;P(\text{8 on time}) \:=\:\begin{pmatrix}8\\8\end{pmatrix}(0.65)^8

Their sum is some decimal p\displaystyle p . . . I'll let you crank it out.

Therefore: \(\displaystyle \,P(\text{at least 6 on time}) \:=\:p\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

(b) During any hour: \(\displaystyle \,P(\text{6 or more on time}) \:=\:p\)

       \displaystyle \;\;\;\,During any hour: P(5 or less on time)=1p=q\displaystyle \,P(\text{5 or less on time}) \:=\:1\,-\,p \:=\:q

For 9 of the next 10 hours, the probability is: \(\displaystyle \L\,\begin{pmatrix}10\\9\end{pmatrix}p^9\,q\)

 
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