Question

4) It is late on Halloween and the candy bowl contains five Almond Joy bars and eight Snickers bars. If you choose four barsb) The probability that at most one is an Almond Joy bar.

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Answer #1

aP(exactly 3 are Almond Joy bars)=P(choose 3 out of 5 Almond Joy bars and 1 out of 8 Snickers bars)

=(5C3)*(8C1)/(13C4)

=10*8/715
=16/143

b)

probability that at most one is an Almond Joy bar =P(X<=1) =P(X=0)+P(X=1)

=(5C0)*(8C4)/(13C4)+(5C1)*(8C3)/(13C4)

=(1*70/715)+(5*56/715)

=70/143

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