Question

If you have 127 different drink and one cup fits 3 drink Q1) how many cups...

If you have 127 different drink and one cup fits 3 drink

Q1) how many cups can you have?

Q2) What is the probability of my friend and I getting the same cup with three drink?

Q3) If there is 20 special drink out of the 127, what is the probability of getting a) cup contains three special drink b) cup contains at least one special drink ?

Math   Statistics-And-Probability   
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Homework Answers

Answer #1

#1.
3 drinks for a cup can be selected in 127C3 = 333375

hence there could be 333375 cups

#2.
Probability of getting a particular cup is 1/333375

Probability of getting a particular cup for you and your friend is 1/333375 * 1 = 1/333375

#3.
a)
cup with 3 special drinks can be formed in 20C3 = 1140 ways
Required proability = 1140/333375 = 0.00342

b)
consider an event with no special drink, i.e. from 107 drinks
This can be formed in 107C3 = 198485 ways

required probability = 1 - (198485/333375) = 0.4046

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