b) The probability that at most one is an Almond Joy bar.
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
b) The probability that at most one is an Almond Joy bar. 4) It is late...
There are 5 Snickers, 10 Baby Ruths, 13 Milky Ways, 12 Twix’s’ and 17 Almond Joys in a bowl of candy. You reach into the bowl and randomly select 5 candy bars. Use this information to answer the following two questions. 1. What is the probability that you select exactly 2 Milky Way bars? 2. What is the probability that you select 3 or fewer Almond Joys? Hints: a) Determine what kind of probability density function this is. Cite 2...