A bowl of Halloween candy has 17 Kit-Kats, 23 Hershey bars, 11 Starbursts, and 14 Skittles packets.
Assuming sampling with replacement
(a) What is the probability that you will select a piece of chocolate candy (Kit Kat or Hershey)?
(b) What is the probability that you will select a Starburst?
Now assuming sampling without replacement:
(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 1 Kit-Kat, 1 Skittles, and 1 Hershey bar, what is the probability that the final candy will be a Starburst?

a.) 8/13 = 0.6154.

b) 11/65 = 0.1692

c.) 0.0267

d.) 0.0222

Step-by-step explanation:

firstly, we determine total number of candies.

17 kit kat(K)+ 23 hershey(H) + 11 starburst(B) + 14 skittles(S) = 65 candies.

Since, Probability = favourable outcome / possible outcome.

With replacement,

a.) probability of chocolate candy (kitkat or hershey) = 17/65 or 23/65

= 17/65 + 23/65

=40/65 = 8/13

b.) probability of selecting a starburst = 11/65

When without replacement:,

c) 2 kit kats, 1 skittles as first three picks and one chocolate candy as the last pick, = [KKSK], [KSKK], [SKKK], [KKSH], [KSKH], [SKKH]

[(17/65 * 16/64 * 14/63 * 15/62) × 3] + [(17/65 * 16/64 * 14/63 * 23/62) ×3]

Note: first bracket and fourth brackets are multiplied by 3 because the value of the first bracket is same for the first three brackets and the second three brackets also have the same values.

Re:chocolate candy means either KitKat or Hershey.

=[(51720/16248960) × 3] + [(87584/16248960) ×3]

=0.0105 +0.0162

= 0.0267.

d.) 1 kitkat, 1 skittles and 1 hershey as first 3 picks and one Starburst as the last pick.

= [KSHB], [KHSB], [SKHB], [SHKB], [HKSB], [HSKB]

= [(17/65 * 14/64 * 23/63 * 11/62) × 6]

note: multiplied by 6 because we will have the same value for all 6 brackets

= 60214/16248960 * 6

= 0.0222.