Question

Candy. Someone hands you a box of a dozen chocolate-covered candies, telling you that half are vanilla creams and the other half peanut butter. You pick candies at random and discover the first three you eat are all vanilla.


a) If there really were 6 vanilla and 6 peanut butter candies in the box, what is the probability that you would have picked three vanillas in a row?


b) Do you think there really might have been 6 of each? Explain.


c) Would you continue to believe that half are vanilla if the fourth one you try is also vanilla? Explain.

Answer 1

Answer:

a) P=0.091

b) If there are half of each taste, picking 3 vainilla in a row has a rather improbable chance (9%), but it is still possible that there are 6 of each taste.

c) The probability of picking 4 vainilla in a row, if there are half of each taste, is P=0.030.

This is a very improbable case, so if this happens we would have reasons to think that there are more than half vainilla candies in the box.

Step-by-step explanation:

We can model this problem with the variable x: number of picked vainilla in a row, following a hypergeometric distribution:

$P(x=k)=\dfrac{\binom{K}{k}\cdot \binom{N-K}{n-k}}{\binom{N}{n}}$

being:

N is the population size (12 candies),

K is the number of success states in the population (6 vainilla candies),

n is the number of draws (3 in point a, 4 in point c),

k is the number of observed successes (3 in point a, 4 in point c),

a) We can calculate this as:

$P(x=3)=\dfrac{\binom{6}{3}\cdot \binom{12-6}{3-3}}{\binom{12}{3}}=\dfrac{\binom{6}{3}\cdot \binom{6}{0}}{\binom{12}{3}}=\dfrac{20\cdot 1}{220}=0.091$

b) If there are half of each taste, picking 3 vainilla in a row has a rather improbable chance (9%), but is possible.

c) In the case k=4, we have:

$P(x=3)=\dfrac{\binom{6}{4}\cdot \binom{6}{0}}{\binom{12}{4}}=\dfrac{15\cdot 1}{495}=0.030$

This is a very improbable case, so we would have reasons to think that there are more than half vainilla candies in the box.

Answer 2

Using the hypergeometric distribution, it is found that:

• a) 0.0909 = 9.09% probability that you would have picked three vanillas in a row.

• b) The probability is above 5%, hence it is not an unusual event and gives no evidence that there might not have been 6 of each.

• c) The probability is below 5%, hence it is an unusual event and there is enough evidence to believe that there might not have been 6 of each.

The candies are chosen without replacement, hence the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• N is the size of the population.

• n is the size of the sample.

• k is the total number of desired outcomes.

Item a:

• There is a total of 12 candies, hence $N = 12$.

• 6 of those candies are vanillas, hence $k = 6$.

• 3 candies are chosen, hence $n = 3$.

The probability that you would have picked three vanillas in a row is P(X = 3), hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 3) = h(3,12,3,6) = \frac{C_{6,3}C_{6,0}}{C_{12,3}} = 0.0909$

0.0909 = 9.09% probability that you would have picked three vanillas in a row.

Item b:

The probability is above 5%, hence it is not an unusual event and gives no evidence that there might not have been 6 of each.

Item c:

Now n = 4, and the probability is P(X = 4), hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 4) = h(4,12,4,6) = \frac{C_{6,4}C_{6,0}}{C_{12,4}} = 0.0303$

The probability is below 5%, hence it is an unusual event and there is enough evidence to believe that there might not have been 6 of each.

To learn more about the hypergeometric distribution, you can take a look at https://brainly.com/question/4818951