Bags of \"Original Skittles\" candies come in five different flavors: Grape, Lem

Question

Bags of "Original Skittles" candies come in five different flavors: Grape, Lemon, Green Apple, Orange and Strawberry. The flavors are equiprobable; if you take a candy randomly from a bag the probability of getting a particular flavor is 1/5, independent of other "draws". (a) What's the probability of choosing three Skittles and getting three of the same flavor? (b) What's the probability of choosing five Skittles and getting one of each flavor? (c) What's the probability that in a bag of 30 candies there aren't any Lemon ones? (d) if you draw a sequence of candies from the bag, what's the probability that you'll get at least two Lemon ones before you get your first Grape one?

Explanation / Answer

(a) Here,

Pr(of each flavour) = 1/5

Pr(choosing three skittlies and getting three of the same flavour) = 5C1 * (1/5)3 = 1/52 = 1/25

Here 5C1 means any 1 out of 5 skittlies and we will use cube as each skittles have 1/5 probability.

(b) Here we have choosen 5 skittles and get 1 each from each flavour.

Now we can see this question as there are 5 places for these skittles. So, at first any type of skittles can come that makes probability 5/5. Now, this type of skittles cannot come at second place so that makes its probability 4/5

so,

Pr(get 1 from each flavour) = 5/5 * 4/5 * 3/5 * 2/5 * 1/5 = 24/625

(c) Here there are 30 candies in the bag. It is like a binomial distribution where n = 30 and p = 1/5

so, Pr( there are no lemon ones) = 30C0 (1/5)0 (4/5)30 = 0.0012

(d) Here are the possibilities are

(i) 2 lemon candies first then first grape one

(ii) 1 lemon candy then first grape one and then the lemon one

(iii) 1 lemon candy first and then two grape one.

as each and every candy has equal probability to come then these three events will also have equal probability.

so Pr(2 lemon ones before first grape one) = 1/3

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