Questions: Mr. Lunce keeps a jar of candy on his desk for students to pick from when they do well on a test. The jar contains 5 Snickers, 2 Butterfingers, 4 Almond Joys, and 3 Milky Ways. If two students get to pick candy from the jar, what is the probability that the first student picks a Snickers and then the second student also picks a Snickers?

Solution Steps

To solve this problem, we need to calculate the probability of two sequential events: the first student picking a Snickers and the second student also picking a Snickers. We will use the concept of conditional probability. First, we calculate the probability of the first student picking a Snickers. Then, assuming the first student picked a Snickers, we calculate the probability of the second student picking a Snickers from the remaining candies.

The total number of candies in the jar is calculated by summing up all the candies: \[ \text{Total candies} = 5 + 2 + 4 + 3 = 14 \]

The probability that the first student picks a Snickers is given by the ratio of Snickers to the total number of candies: \[ P(\text{First Snickers}) = \frac{5}{14} \approx 0.3571 \]

After the first student picks a Snickers, the number of Snickers and the total number of candies are reduced by one: \[ \text{Remaining Snickers} = 5 - 1 = 4 \] \[ \text{Remaining candies} = 14 - 1 = 13 \]

The probability that the second student picks a Snickers is given by the ratio of the remaining Snickers to the remaining candies: \[ P(\text{Second Snickers}) = \frac{4}{13} \approx 0.3077 \]

The total probability that both the first and second students pick a Snickers is the product of the two probabilities: \[ P(\text{Both Snickers}) = P(\text{First Snickers}) \times P(\text{Second Snickers}) = 0.3571 \times 0.3077 \approx 0.1099 \]

Final Answer