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zhenek [66]
3 years ago
8

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
Mathematics
1 answer:
e-lub [12.9K]3 years ago
7 0

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}

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