Ask Question
14 April, 07:13

Howard chose a candy from a bowl with 5 chocolate candies, 4 gummy candies and 6 hard candies. What is Howard's dependent probability of choosing a chocolate candy followed by a gummy candy?

Round to the nearest tenth of a percent, if necessary.

+3
Answers (1)
  1. 14 April, 09:56
    0
    8.9%

    Step-by-step explanation:

    Here, we are to calculate the probability of Howard choosing a chocolate candy followed by a gummy candy.

    The probability of selecting a chocolate candy = number if chocolate candy / total number of candy

    Total number of candy = 5 + 4 + 6 = 15

    Number of chocolate candy = 5

    The probability of selecting a chocolate candy = 5/15 = 1/3

    The probability of selecting a gummy candy = number of gummy candies/total number of candies

    Number of gummy candy = 4

    The probability of selecting a gummy candy = 4/15

    The probability of selecting a chocolate candy before a gummy candy = 1/3 * 4/15 = 4/45 = 0.088888888889

    Which is same as 8.89 percent which is 8.9% to the nearest tenth of a percent
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Howard chose a candy from a bowl with 5 chocolate candies, 4 gummy candies and 6 hard candies. What is Howard's dependent probability of ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers