Question

a package contains 4 red, 2 green, 8 purple, and 6 blue jelly beans. what is the probability of choosing a purple jelly bean, eating it, and then choosing a blue jelly bean?
\\(\frac{1}{400}\\)
\\(\frac{1}{380}\\)
\\(\frac{3}{25}\\)
\\(\frac{12}{95}\\)

Explanation:

Step1: Calculate total jelly - beans initially

The total number of jelly - beans initially is $4 + 2+8 + 6=20$.

Step2: Calculate probability of choosing a purple jelly - bean first

The probability of choosing a purple jelly - bean first, $P(purple)=\frac{8}{20}=\frac{2}{5}$.

Step3: Calculate number of jelly - beans after eating a purple one

After eating a purple jelly - bean, the number of jelly - beans left is $20 - 1 = 19$.

Step4: Calculate probability of choosing a blue jelly - bean second

The probability of choosing a blue jelly - bean second, $P(blue)=\frac{6}{19}$.

Step5: Calculate the combined probability

Since these are independent events (the first event affects the sample space for the second event), the probability of choosing a purple jelly - bean first and then a blue jelly - bean is $P = P(purple)\times P(blue)=\frac{2}{5}\times\frac{6}{19}=\frac{12}{95}$.

Answer:

$\frac{12}{95}$