Question:
There are 30 candies in a box, all identically shaped. 5 are filled with coconut, 10 with caramel, and 15 are solid chocolate.
You randomly select a piece of candy and eat it (so it is NOT replaced!), then select a second piece. Find the probability of each event
(a) The probability of selecting two solid chocolates in a row.
(b) The probability of selecting a caramel and then a coconut candy.
Answer:
[tex](a)[/tex] [tex]P(Chocolates) = \frac{7}{29}[/tex]
[tex](b)[/tex] [tex]P(Caramel\ and\ Coconut) = \frac{5}{87}[/tex]
Step-by-step explanation:
Given
[tex]Coconut = 5[/tex]
[tex]Caramel = 10[/tex]
[tex]Chocolate = 15[/tex]
[tex]Total = 30[/tex]
For probabilities without replacement, 1 is subtracted after the first selection.
So, we have:
Solving (a): Two solid chocolates
This is calculated as:
[tex]P(Chocolates) = P(First\ Chocolate) * P(Second\ Chocolate)[/tex]
[tex]P(Chocolates) = \frac{n(Chocolate)}{Total} * \frac{n(Chocolate) - 1}{Total - 1}[/tex]
[tex]P(Chocolates) = \frac{15}{30} * \frac{15 - 1}{30 - 1}[/tex]
[tex]P(Chocolates) = \frac{15}{30} * \frac{14}{29}[/tex]
[tex]P(Chocolates) = \frac{1}{2} * \frac{14}{29}[/tex]
[tex]P(Chocolates) = \frac{7}{29}[/tex]
Solving (a): Caramel and Coconut
This is calculated as:
[tex]P(Caramel\ and\ Coconut) = P(Caramel) * P(Coconut)[/tex]
[tex]P(Caramel\ and\ Coconut) = \frac{n(Caramel)}{Total} * \frac{n(Coconut)}{Total - 1}[/tex]
[tex]P(Caramel\ and\ Coconut) = \frac{10}{30} * \frac{5}{30- 1}[/tex]
[tex]P(Caramel\ and\ Coconut) = \frac{10}{30} * \frac{5}{29}[/tex]
[tex]P(Caramel\ and\ Coconut) = \frac{1}{3} * \frac{5}{29}[/tex]
[tex]P(Caramel\ and\ Coconut) = \frac{5}{87}[/tex]
