Question

selects a piece of candy and eats it (so it is NOT replaced!) Then selects a piece of candy and eats it. Find the probability of each event

Answer 1

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:

$(a)$ $P(Chocolates) = \frac{7}{29}$

$(b)$ $P(Caramel\ and\ Coconut) = \frac{5}{87}$

Step-by-step explanation:

Given

$Coconut = 5$

$Caramel = 10$

$Chocolate = 15$

$Total = 30$

For probabilities without replacement, 1 is subtracted after the first selection.

So, we have:

Solving (a): Two solid chocolates

This is calculated as:

$P(Chocolates) = P(First\ Chocolate) * P(Second\ Chocolate)$

$P(Chocolates) = \frac{n(Chocolate)}{Total} * \frac{n(Chocolate) - 1}{Total - 1}$

$P(Chocolates) = \frac{15}{30} * \frac{15 - 1}{30 - 1}$

$P(Chocolates) = \frac{15}{30} * \frac{14}{29}$

$P(Chocolates) = \frac{1}{2} * \frac{14}{29}$

$P(Chocolates) = \frac{7}{29}$

Solving (a): Caramel and Coconut

This is calculated as:

$P(Caramel\ and\ Coconut) = P(Caramel) * P(Coconut)$

$P(Caramel\ and\ Coconut) = \frac{n(Caramel)}{Total} * \frac{n(Coconut)}{Total - 1}$

$P(Caramel\ and\ Coconut) = \frac{10}{30} * \frac{5}{30- 1}$

$P(Caramel\ and\ Coconut) = \frac{10}{30} * \frac{5}{29}$

$P(Caramel\ and\ Coconut) = \frac{1}{3} * \frac{5}{29}$

$P(Caramel\ and\ Coconut) = \frac{5}{87}$