# Choosing a distribution

## Main Question or Discussion Point

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 probabilty 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 probabilty 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!