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 February 25th, 2017, 07:10 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, 07:37 PM   #2
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Quote:
 Originally Posted by palongze 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]
This is just the binomial probability distribution.

$p=\dfrac {1}{20}$

$P[\text{k late out of n deliveries}] = \binom{n}{k}p^k (1-p)^{n-k}$

a) $P[\text{at least 1 late delivery}]=1-P[\text{no late deliveries}]$

$P[\text{no late deliveries}] = (1-p)^n$

$P[\text{at least 1 late delivery}]=1 - (1-p)^n < 0.8$

This is easily solved to obtain $n=31$

b) $(1-p)^n > 0.1$

again easily solved to obtain $n=44$

Last edited by romsek; February 25th, 2017 at 08:32 PM.

 February 25th, 2017, 07:43 PM #3 Senior Member   Joined: May 2016 From: USA Posts: 888 Thanks: 355 $a = P(at\ least\ one\ late\ delivery) < 0.8.$ $\therefore b = P(no\ late\ delivery) = 1 - a \implies b > 0.2.$ $b = \dbinom{n}{0} * 0.95^n * 0.05^{(n - n)} = 0.95^n.$ $0.95^{max(n)} > 0.2 \implies max(n)* ln(0.95) > ln(0.2) \implies max(n) > \dfrac{ln(0.2)}{ln(0.95)} > 31 \implies$ $max(n) = 31.$ 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? Thanks from palongze
February 25th, 2017, 08:56 PM   #4
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Quote:
 Originally Posted by romsek This is just the binomial probability distribution. $p=\dfrac {1}{20}$ $P[\text{k late out of n deliveries}] = \binom{n}{k}p^k (1-p)^{n-k}$ a) $P[\text{at least 1 late delivery}]=1-P[\text{no late deliveries}]$ $P[\text{no late deliveries}] = (1-p)^n$ $P[\text{at least 1 late delivery}]=1 - (1-p)^n < 0.8$ This is easily solved to obtain $n=31$ b) $(1-p)^n > 0.1$ again easily solved to obtain $n=44$
Thank you. I understand where my mistake now.

Quote:
 Originally Posted by JeffM1 $a = P(at\ least\ one\ late\ delivery) < 0.8.$ $\therefore b = P(no\ late\ delivery) = 1 - a \implies b > 0.2.$ $b = \dbinom{n}{0} * 0.95^n * 0.05^{(n - n)} = 0.95^n.$ $0.95^{max(n)} > 0.2 \implies max(n)* ln(0.95) > ln(0.2) \implies max(n) > \dfrac{ln(0.2)}{ln(0.95)} > 31 \implies$ $max(n) = 31.$ 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?
Yes, I do understand your working step. Thank you mate.

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