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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?

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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: [tex]\dfrac{10}{91}[/tex]

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 =[tex]P(S_1)=\dfrac{5}{14}[/tex]

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 = [tex]P(S_2|S_1)\dfrac{4}{13}[/tex]

We know that [tex]P(A\cap B)=P(B|A)\times P(A)[/tex]

Thus, [tex]P(S_1\cap S_2)=P(S_2|S_1)\times P(S_1)[/tex]

[tex]=\dfrac{4}{13}\times\dfrac{5}{14}=\dfrac{10}{91}[/tex]

Hence, the probability that the first student picks a snickers and then the second student also picks a snickers [tex]=\dfrac{10}{91}[/tex]

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