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 July 2nd, 2017, 08:17 PM #1 Newbie   Joined: Jul 2017 From: Melbourne Posts: 1 Thanks: 0 Is this problem solvable or deliberately absurd? For context, I am an actor and have been given this maths problem to say in a script (I'm playing a teacher) and for the life of me I can't make sense of it. Its possible that it is deliberately nonsensical but I would like to make sure. "Monique is practicing her shopping for netball. She knows from past experience that the probability of her making any one shot is 70%. Her coach has asked her to keep practicing until she scores 50 goals. How many shots would she need to attempt to ensure that the probability of making at least 50 shots is more than 0.99?" The answer is later given at 86, but I would to be able to make sense of the mechanics of it.
July 2nd, 2017, 08:21 PM   #2
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Quote:
 Originally Posted by Leturtle Monique is practicing her shopping for netball.
What the heck does that mean?!

July 2nd, 2017, 08:35 PM   #3
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Quote:
 Originally Posted by Denis What the heck does that mean?!
The art of selecting the right (net?)ball.

July 2nd, 2017, 08:51 PM   #4
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Quote:
 Originally Posted by Leturtle Monique is practicing her shopping for netball.
Shooting, obviously.

$\Pr{(score)}= \frac7{10}$

Quote:
 Originally Posted by Leturtle How many shots would she need to attempt to ensure that the probability of making at least 50 shots is more than 0.99?
So we need to find $n$ such that $\Pr{(\text{scoring$50$or more shots out of$n$})} \gt \frac{99}{100}$.

Normally we might use a binomial distribution, but here there are rather a lot of trials (shots) to be taken and we'd need to deal with 50, 51, 52, ... scores. I would suggest using an appropriate approximation to the Binomial distribution.

 July 3rd, 2017, 02:20 AM #5 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics According to the approximation method in R (which uses the beta distribution), 87 is the answer: pbinom(50, 86, 0.7, FALSE) # returns 0.9871295 pbinom(50, 87, 0.7, FALSE) # returns 0.9911758 That's close enough to the correct answer! With smaller numbers, this would be a pretty 'standard' question in elementary probability. Edit: The exact probability when n = 86 is 0.9911758. When n = 85, it's 0.9895492. This verifies the answer. sum = 0 for(i in 50:86){ sum = sum + choose(86,i) * .7^i * .3^(86-i)} #Returns 0.9911758 sum = 0 for(i in 50:85){ sum = sum + choose(85,i) * .7^i * .3^(85-i)} #Returns 0.9895492 Last edited by 123qwerty; July 3rd, 2017 at 02:35 AM.
 July 3rd, 2017, 05:13 PM #6 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 580 Thanks: 80 http://stattrek.com/online-calculator/binomial.aspx agrees that 86 is the answer. Last edited by EvanJ; July 3rd, 2017 at 05:15 PM.

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