Fruit Sample

[sportywarbz]

A customer from Cavallaro's Fruit Stand picks a sample of 3 oranges at random from a crate containing 65 oranges, of which 4 are rotten. What is the probability that the sample contains 1 or more rotten oranges? (Round your answer to three decimal places.)

This is what I have figured so far...I don't know if its right. Divide this up into rotten and not rotten oranges. The denominator will be the number of ways of picking any 3 from the entire group of oranges (not separated by rotten/not rotten). So like (3/65)(2/64)(1/63) or 65 C 3?

 

[soroban]

Hello, sportywarbz!

A customer from Cavallaro's Fruit Stand picks a sample of 3 oranges at random from a crate containing 65 oranges, of which 4 are rotten.
What is the probability that the sample contains 1 or more rotten oranges? (Round your answer to three decimal places.)

You are correct . . .
\(\displaystyle \text{The denominator is the number of ways to choose 3 oranges from the enire 65 oranges: }\:_{65}C_3 \:=\:{65\choose3} \:=\:\frac{65!}{3!\,62!} \;=\;43,680\)


The opposite of "one or more rotten oranges" is "no rotten oranges".


To get no rotten oranges, there are: .\(\displaystyle _{61}C_3 \;=\;35,\!990\text{ ways.}\)

. . \(\displaystyle \text{Hence: }\(\text{no rotten oranges}) \:\;=\;\:\frac{35,\!990}{43,\!680} \:\;=\;\:\frac{3599}{4368}\)

\(\displaystyle \text{Therefore: }\(\text{1 or more rotten oranges}) \:\;=\;\:1 \;-\; \frac{3599}{4368} \;\;=\;\;\frac{769}{4368} \;\;=\;\;0.176053114 \;\;\approx\;\;0.176\)