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July 2nd, 2017, 07:17 PM   #1
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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.
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July 2nd, 2017, 07:21 PM   #2
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Monique is practicing her shopping for netball.
What the heck does that mean?!
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July 2nd, 2017, 07:35 PM   #3
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Originally Posted by Denis View Post
What the heck does that mean?!
The art of selecting the right (net?)ball.
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July 2nd, 2017, 07:51 PM   #4
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Quote:
Originally Posted by Leturtle View Post
Monique is practicing her shopping for netball.
Shooting, obviously.

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

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Originally Posted by Leturtle View Post
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.
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July 3rd, 2017, 01:20 AM   #5
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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 01:35 AM.
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July 3rd, 2017, 04:13 PM   #6
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http://stattrek.com/online-calculator/binomial.aspx agrees that 86 is the answer.

Last edited by EvanJ; July 3rd, 2017 at 04:15 PM.
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