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algol [13]
4 years ago
5

Scoring a hole-in-one is the greatest shot a golfer can make. Once 5 professional golfers each made holes-in-one on the 7th hole

at the same golf course at the same tournament. It has been found that the estimated probability of making a hole-in-one is (1/2494) for male professionals. Suppose that a sample of 5 professional male golfers is randomly selected.
(a) What is the probability that at least one of these golfers makes a hole-in-one on the 10th hole at the same tournament?
(b) What is the probability that all of these golfers make a hole-in-one on the 10th hole at the same tournament?
Mathematics
1 answer:
adell [148]4 years ago
5 0
A) Probability  that at least one of these golfers makes a hole in one on the 10th hole at the same tournament is 

<span>=1−(<span>2493/2494</span><span>)^<span>5
=1-(0.99)^5
= 1- 0.96
=0.4

b) T</span></span></span><span>he probability that all of these golfers make a hole-in-one on the 10th hole at the same tournament
=</span>(<span>1/2494</span><span>)<span>5
</span></span>
which is nearly zero.
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Ahh! What a classic: Systems of Equations!

So, if you don't know what a system of equations is, its basically two equations with two variables combined into a system.

So, let x=# of pigs

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Let's make our first equation!

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So then, knowing that a pig has 4 legs and a chicken 2:

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This is because there are 40 legs in total, and we multiply the variable because that's how many legs each creature has.

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Then, we input it into the next equation:

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This means that there are 6 chickens, and by inputting it into the first equation:

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So there you have it!

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Hope this helps!

P.S. Stay Safe!

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