Answer:
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.
