Selecting a Distribution for Orange Juice Taste Testing with N Judges

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SUMMARY

The discussion centers on determining the probability that at least one judge selects brand A in a taste testing experiment involving 10 judges and four brands of orange juice. The probability of a judge choosing brand A is established at 1/4. Brendan correctly identifies the use of the binomial distribution to calculate the probability of at least one judge selecting brand A, concluding that the probability is approximately 0.943 after correcting his initial calculations.

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boneill3
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I have been given a question that reads:
In a taste testing experiment N judges are offered four different brands of orange juice to taste.
Suppose Y is the number of of judges that choose brand A as the best.

I know that the probability of a judge choosing A is 1/4.

I have been asked what is the probability that at least one judge chooses brand A given that there are 10 judges.

My question is what would be the best distribution to use?

As I'm going to have to simulate 1000 trials.

Thanks
Brendan
 
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No need to simulate. Think of it this way... what is the probability that NO judge chooses A?
 
The probability of not A is 0.75 but that is for 1 judge, but the probability of 10 judges not picking juice A would be a lot lower.

I thought I might be able to use a binomial distribution to figure it out so I can simulate for any number of judges?
regards
Brendan
 
Just wondering if this makes sense.

Let x = 1 be the number of successes (here the success is that the judges don't pick A)
Let n = the number of judges = 10

So P (x) = 0.75

If I use the binomial distribution

P(x=0) = 10Choose0 * P(x)^x * P(not x)^n-x

P(x=0) = 10Choose1 * (3/4)^0 * (1/4)^10

P(x=0) = 0.000000953

In another words it would be extremely improbable to have no judge pick A
 
So would the probability of at least one picking juice A be P(1-P(x=0) = 0.000000953)

> 99.9

regards
Brendan
 
Sorry guy's I think I've got it wrong.
I'll try again!

Let x = 1 be the number of successes (here the success is that the judges pick A)
Let n = the number of judges = 10

So P (x) = 0.25

If I use the binomial distribution to find the probability of 10 judges not choosing A

P(x=0) = 10Choose0 * P(x)^x * P(not x)^n-x

P(x=0) = 10Chooseo * (1/4)^0 * (3/4)^10

P(x=0) = 0.056

So the probability of at least 1 must be (1 - 0.056) = 0.943

I hope this is right cause I'm loosing my mind!
Brendan
 
Yep, 0.943 is right.
 
Thanks mate!
 

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