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Chapter 1: Problem 44

Urn 1 has five white and seven black balls. Urn 2 has three white and twelve black balls. We flip a fair coin. If the outcome is heads, then a ball from urn 1 is selected, while if the outcome is tails, then a ball from urn 2 is selected. Suppose that a white ball is selected. What is the probability that the coin landed tails?

Short Answer

Expert verified
The probability that the coin landed tails, given that a white ball was selected, is \(\frac{3}{13}\), or approximately 0.231.

Step by step solution

01

Identify given probabilities

: First, let's denote the events: - H: coin lands heads - T: coin lands tails - W: white ball is selected - B: black ball is selected We have the following probabilities: 1. \(P(H) = 0.5\) (probability of heads) 2. \(P(T) = 0.5\) (probability of tails) 3. \(P(W|H) = \frac{5}{12}\) (probability of a white ball given heads) 4. \(P(W|T) = \frac{3}{15}\) (probability of a white ball given tails)
02

Use Bayes' theorem

: We want to calculate \(P(T|W)\), the probability that the coin landed tails given that a white ball was selected. Using Bayes' theorem, we have: \[P(T|W) = \frac{P(W|T) \cdot P(T)}{P(W|T) \cdot P(T) + P(W|H) \cdot P(H)}\]
03

Substitute probabilities

: Now we substitute the probabilities we have gathered in step 1: \[\begin{aligned} P(T|W) &= \frac{\frac{3}{15} \cdot 0.5}{\frac{3}{15} \cdot 0.5 + \frac{5}{12} \cdot 0.5}\\ &= \frac{ \frac{3}{30}}{\frac{3}{30} + \frac{5}{24}} \end{aligned}\]
04

Calculate

: Now we can calculate the probability of the coin landing tails given that a white ball was selected: \[P(T|W) = \frac{3}{3 + 10} = \frac{3}{13}\]
05

Interpret the result

: The probability that the coin landed tails, given that a white ball was selected, is \(\frac{3}{13}\), or approximately 0.231.

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Most popular questions from this chapter

Suppose each of three persons tosses a coin. If the outcome of one of the tosses differs from the other outcomes, then the game ends. If not, then the persons start over and retoss their coins. Assuming fair coins, what is the probability that the game will end with the first round of tosses? If all three coins are biased and have probability \(\frac{1}{4}\) of landing heads, what is the probability that the game will end at the first round?

Show that $$ P\left(\bigcup_{i=1}^{n} E_{i}\right) \leqslant \sum_{i=1}^{n} P\left(E_{i}\right) $$ This is known as Boole's inequality. Hint: Either use Equation (1.2) and mathematical induction, or else show that \(\bigcup_{i=1}^{n} E_{i}=\bigcup_{i=1}^{n} F_{i}\), where \(F_{1}=E_{1}, F_{i}=E_{i} \bigcap_{j=1}^{i-1} E_{j}^{c}\), and use property (iii) of a probability.

(a) A gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random, and when he flips it, it shows heads. What is the probability that it is the fair coin? (b) Suppose that he flips the same coin a second time and again it shows heads. Now what is the probability that it is the fair coin? (c) Suppose that he flips the same coin a third time and it shows tails. Now what is the probability that it is the fair coin?

If \(P(E)=0.9\) and \(P(F)=0.8\), show that \(P(E F) \geqslant 0.7\). In general, show that $$ P(E F) \geqslant P(E)+P(F)-1 $$ This is known as Bonferroni's inequality.

A box contains three marbles: one red, one green, and one blue. Consider an experiment that consists of taking one marble from the box then replacing it in the box and drawing a second marble from the box. What is the sample space? If, at all times, each marble in the box is equally likely to be selected, what is the probability of each point in the sample space?
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