Question

consider two unopened bags of candy (one skittles, the other m&ms). the manufacturers state that the distribution of coloured candies is: m&ms: 23% blue, 20% orange, 15% green, 14% yellow, 13% red, 15% brown. skittles: colours are distributed evenly, meaning each colour (red, orange, green, blue, yellow, and purple) has a probability of 1/5. assume that both of these unopened bags have the same number of candies. the bags are opened and the candies are mixed in a bowl. if one orange candy is randomly selected from this bowl, what is the probability it is a skittle? (hint: since both unopened bags have the same number of candies, the probability of choosing a(n) orange m&m from the mixed bowl is half of the probability of choosing a(n) orange m&m from the unopened bag of m&ms, and similarly for the probability of choosing a(n) orange skittle from the mixed bowl.) p(skittle | orange)=0.4547 (if rounding, use at least four digits after the decimal in your answer)

Explanation:

Step1: Calculate probability of orange M&M in mixed - bowl

Since the probability of an orange M&M in an un - opened bag is 20% or 0.2, and in the mixed bowl (with equal number of candies from each bag), the probability of an orange M&M, $P(\text{orange M&M})=\frac{0.2}{2}=0.1$.

Step2: Calculate probability of orange Skittle in mixed - bowl

Since the probability of an orange Skittle in an un - opened bag is $\frac{1}{5}=0.2$, and in the mixed bowl, the probability of an orange Skittle, $P(\text{orange Skittle})=\frac{0.2}{2}=0.1$.

Step3: Use Bayes' theorem formula for conditional probability

The formula for $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In the case of $P(\text{Skittle}|\text{orange})=\frac{P(\text{orange Skittle})}{P(\text{orange Skittle}) + P(\text{orange M&M})}$.
Substitute the values: $P(\text{Skittle}|\text{orange})=\frac{0.1}{0.1 + 0.1}=\frac{0.1}{0.2}=0.5$.

Answer:

0.5