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February 25th, 2017, 06:10 PM  #1 
Newbie Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus  Probability finding "n"
Hi everyone. I have problem regarding solving this questions. Hope some one can help. Q: For a new express delivery company, the probability of making a late delivery per day is one in twenty. Calculate the maximum number of deliveries per day if the probability of (a) at least one late delivery is to be less than 0.8 (b) no late delivery is to be more than 0.1 [ans: a=31, b=44] 
February 25th, 2017, 06:37 PM  #2  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,397 Thanks: 709  Quote:
$p=\dfrac {1}{20}$ $P[\text{k late out of n deliveries}] = \binom{n}{k}p^k (1p)^{nk}$ a) $P[\text{at least 1 late delivery}]=1P[\text{no late deliveries}]$ $P[\text{no late deliveries}] = (1p)^n$ $P[\text{at least 1 late delivery}]=1  (1p)^n < 0.8$ This is easily solved to obtain $n=31$ b) $(1p)^n > 0.1$ again easily solved to obtain $n=44$ Last edited by romsek; February 25th, 2017 at 07:32 PM.  
February 25th, 2017, 06:43 PM  #3 
Senior Member Joined: May 2016 From: USA Posts: 785 Thanks: 311 
$a = P(at\ least\ one\ late\ delivery) < 0.8.$ $\therefore b = P(no\ late\ delivery) = 1  a \implies b > 0.2.$ $b = \dbinom{n}{0} * 0.95^n * 0.05^{(n  n)} = 0.95^n.$ $0.95^{max(n)} > 0.2 \implies max(n)* ln(0.95) > ln(0.2) \implies max(n) > \dfrac{ln(0.2)}{ln(0.95)} > 31 \implies$ $max(n) = 31.$ Let's check that answer. $1  0.95^{31} \approx 0.796 < 0.8,\ but\ 1  0.95^{32} \approx 0.806 > .8.$ Do you follow that? 
February 25th, 2017, 07:56 PM  #4  
Newbie Joined: Jan 2017 From: Malaysia Posts: 20 Thanks: 3 Math Focus: calculus  Quote:
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