One.
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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:
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:
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:
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:
Step-by-step explanation:
