Question

The popular candy Skittles comes in 5 colors. According to the Skittles website, the 5 colors are evenly distributed in the population of Skittle candies. So each color makes up 20% of the population. Suppose that we purchase a small bag of Skittles. Assume this size bag always has 40 candies. In this particular bag 10 are green. What is the probability that a randomly selected bag of this size has 10 or more green candies

Answer 1

`P(x\geq 10)=0.2682`

Explanation:

The number x of green candies in a bag of 40 candies follows a binomial distribution, because we have:

• n identical and independent events: 40 candies
• a probability p of success and (1-p) of fail: a probability of 0.2 to get a green candie and 0.8 to doesn't get a green candie.

So, the probability that in a bag of 40 candies, x are green is calculated as:

`P(x)=(n!)/(x!(n-x)!)*p^(x)*(1-p)^(n-x)`

Replacing, n by 40 and p by 0.2, we get:

`P(x)=(40!)/(x!(40-x)!)*0.2^(x)*(1-0.2)^(40-x)`

So, the probability that a randomly selected bag of this size has 10 or more green candies is equal to:

`P(x\geq 10)=P(10)+P(11)+...+P(40)\\P(x\geq 10)=1-P(x<10)`

Where
`P(x<10)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)`

So, we can calculated P(0) and P(1) as:

`P(0)=(40!)/(0!(40-0)!)*0.2^(0)*(1-0.2)^(40-0)=0.00013\\P(1)=(40!)/(1!(40-1)!)*0.2^(1)*(1-0.2)^(40-1)=0.00133`

At the same way, we can calculated P(2), P(3), P(4), P(5), P(6), P(7), P(8) and P(9) and get that P(x<10) is equal to:

`P(x<10)=0.7318`

Finally, the probability
`P(x\geq 10)` that a randomly selected bag of this size has 10 or more green candies is:

`P(x\geq 10)=1-P(x<10)\\P(x\geq 10)=1-0.7318\\P(x\geq 10)=0.2682`

Answer 2

27.76% probability that a randomly selected bag of this size has 10 or more green candies

Explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

In this problem, we have that:

`n = 40, p = 0.2`

So

`\mu = E(X) = np = 40*0.2 = 8`

`\sigma = √(V(X)) = √(np(1-p)) = √(40*0.2*0.8) = 2.53`

What is the probability that a randomly selected bag of this size has 10 or more green candies

Using continuity correction, this is
`P(X \geq 10 - 0.5) = P(X \geq 9.5)`, which is 1 subtracted by the pvalue of Z when X = 9.5. So

`Z = (X - \mu)/(\sigma)`

`Z = (9.5 - 8)/(2.53)`

`Z = 0.59`

`Z = 0.59` has a pvalue of 0.7224

1 - 0.7224 = 0.2776

27.76% probability that a randomly selected bag of this size has 10 or more green candies