A bowl of Halloween candy has 17 Kit-Kats, 23 Hershey bars, 11 Starbursts, and l
ID: 3181704 • Letter: A
Question
A bowl of Halloween candy has 17 Kit-Kats, 23 Hershey bars, 11 Starbursts, and l4 Skittles packets. a.) What is the probability that you will select a piece of chocolate candy? b.) What is the probability that you will select a Starburst? c.) If you select a random sample of n=4 and the first three candies are a 2 Kit-Kats and a skittles, what is the probability that you will select a chocolate piece of candy as the fourth item? d.) If you select a random sample of n = 4 and the first three candies are l Kit-Kat, l skittles, and Hershey bar, what is the probability that the final candy will be a Starburst?Explanation / Answer
Total Halloween candy = 17+23+11+14=65
a) P(a chocolate candy)=23/65
Because there are 23 Heeshey bars out of 65 candies.
b) P(starbust) = 11/65 (Same logic as above)
c) n = 4
So total number of ways we can draw a random sample of n=4 is
P=(17*16*14*23)/(65*64*63*62)
Denominator-(Because at first you can select 1 out of 65 candies then you have to choose any one out of remaining 64 candies and then 63 and 62.)
Numerator- similarly you can select any one kitkat out of 17 kitkats then from the remaining 16 kitkats, then one out of 14 skittles, and then one out of 23 Hershey bars.
d) p=(17*14*23*11)/(65*64*63*62) (same logic as above)
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