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Mr. lance 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?

Verified Answer :

texaschic101

there are 5 snickers ...and a total of 14 candy bars

probability of first kid getting a snickers = 5/14
probability of 2nd kid getting a snickers = 4/13

probability of both = 5/14 * 4/13 = 20/182 reduces to 10/91

JeanaShupp

Answer: \(\dfrac{10}{91}\)

Step-by-step explanation:

Given : The jar contains 5 snickers, 2 butterfingers, 4 almond joys and 3 milky ways.

Total candies in the jar = 5+2+4+3=14

Probability of drawing first candy a snickers =\(P(S_1)=\dfrac{5}{14}\)

After this, number of candies left in the jar = 14-1=13

Number of snickers left = 5-1=4

Then the conditional probability of getting another snickers = \(P(S_2|S_1)\dfrac{4}{13}\)

We know that \(P(A\cap B)=P(B|A)\times P(A)\)

Thus, \(P(S_1\cap S_2)=P(S_2|S_1)\times P(S_1)\)

\(=\dfrac{4}{13}\times\dfrac{5}{14}=\dfrac{10}{91}\)

Hence, the probability that the first student picks a snickers and then the second student also picks a snickers \(=\dfrac{10}{91}\)

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