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Chapter 11: Problem 36

We return to our box of chocolates. There are 30 chocolates in the box, all identically shaped. Five are filled with coconut, 10 with caramel, and 15 are solid chocolate. You randomly select one piece, eat it, and then select a second piece. Find the probability of selecting a coconut-filled chocolate followed by a solid chocolate.

Short Answer

Expert verified
The probability of first picking a coconut-filled chocolate and then a solid chocolate is \( \frac{1}{6} \times \frac{15}{29}\)

Step by step solution

01

Compute the probability of first event

This is the probability of picking a coconut chocolate first. Since there are 5 coconut chocolates and 30 in total, this is computed by dividing the number of favorable outcomes by the total number of outcomes. So, \( P(\text{coconut first}) = \frac{5}{30} = \frac{1}{6} \).
02

Compute the probability of second event after the first has occurred

This is the probability of picking a solid chocolate after a coconut chocolate has already been picked and eaten. Now, we have 15 solid chocolates and 29 chocolates in total. Therefore, the probability of picking a solid chocolate in the second pick is \( P(\text{solid second} | \text{coconut first}) = \frac{15}{29} \). This is a conditional probability, where the condition is that a coconut chocolate was picked first.
03

Compute the joint probability

Once we have the two individual probabilities, we can compute the joint probability of two events by multiplying their individual probabilities. Thus, the probability of first picking a coconut chocolate and then a plain chocolate is \( P(\text{coconut first and solid second}) = P(\text{coconut first}) \times P(\text{solid second} | \text{coconut first}) = \frac{1}{6} \times \frac{15}{29} \).

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Most popular questions from this chapter

A game is played using one die. If the die is rolled and shows 1 , the player wins \(\$ 5\). If the die shows any number other than 1 , the player wins nothing. If there is a charge of \(\$ 1\) to play the game, what is the game's expected value? What does this value mean?

Nine cards numbered from 1 through 9 are placed into a box and two cards are selected without replacement. Find the probability that both numbers selected are odd, given that their sum is even.

An architect is considering bidding for the design of a new museum. The cost of drawing plans and submitting a model is \(\$ 10,000\). The probability of being awarded the bid is \(0.1\), and anticipated profits are \(\$ 100,000\), resulting in a possible gain of this amount minus the \(\$ 10,000\) cost for plans and a model. What is the expected value in this situation? Describe what this value means.

Make Sense? Determine whether each statement makes sense or does not make sense, and explain your reasoning. I must have made an error calculating probabilities because \(P(A \mid B)\) is not the same as \(P(B \mid A)\).

Involve computing expected values in games of chance. The spinner on a wheel of fortune can land with an equal chance on any one of ten regions. Three regions are red, four are blue, two are yellow, and one is green. A player wins \(\$ 4\) if the spinner stops on red and \(\$ 2\) if it stops on green. The player loses \(\$ 2\) if it stops on blue and \(\$ 3\) if it stops on yellow. What is the expected value? What does this mean if the game is played ten times?
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