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Chapter 3: Problem 38

Urn \(A\) has 5 white and 7 black balls. Urn \(B\) has 3 white and 12 black balls. We flip a fair coin. If the outcome is heads, then a ball from urn \(A\) is selected, whereas if the outcome is tails, then a ball from urn \(B\) 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 \(\frac{12}{85}\).

Step by step solution

01

Probability of selecting a white ball from each urn

Let's first find the probability of selecting a white ball from each urn. We are given the number of white and black balls in each urn: Urn A: 5 white balls, 7 black balls (total 12 balls) Urn B: 3 white balls, 12 black balls (total 15 balls) The probability of selecting a white ball from each urn is: Urn A: \(P(W_A) = \frac{5}{12}\) Urn B: \(P(W_B) = \frac{3}{15} = \frac{1}{5}\)
02

Probability of flipping a fair coin

The probability of flipping a fair coin to get heads (selecting a ball from urn A) or tails (selecting a ball from urn B) is equal: \(P(Heads) = P(A) = \frac{1}{2}\) \(P(Tails) = P(B) = \frac{1}{2}\)
03

Use Bayes' theorem

Now that we have the probabilities of selecting a white ball from each urn and the probabilities of flipping a fair coin, we can use Bayes' theorem to calculate the probability of the coin landing tails given that a white ball was selected. Bayes鈥� theorem states: \(P(B|W) = \frac{P(W|B) P(B)}{P(W)}\) where \(P(B|W)\) = probability of the coin landing tails given that a white ball was selected, \(P(W|B)\) = probability of selecting a white ball from urn B, \(P(B)\) = probability of the coin landing tails, \(P(W)\) = probability of selecting a white ball from any urn.
04

Calculate the probability of selecting a white ball from any urn

To calculate the probability of selecting a white ball from any urn, we can use the law of total probability: \(P(W) = P(W_A)P(A) + P(W_B)P(B)\) Plugging in the values we found in steps 1 and 2, this becomes: \(P(W) = \frac{5}{12} \times \frac{1}{2} + \frac{1}{5} \times \frac{1}{2} = \frac{5}{24} + \frac{1}{10}\) To add the fractions, we need to find a common denominator: \(P(W) = \frac{5}{24} + \frac{12}{24} = \frac{17}{24}\)
05

Apply Bayes' theorem

Now we have all the values needed to apply Bayes' theorem: \(P(B|W) = \frac{P(W_B)P(B)}{P(W)} = \frac{(\frac{1}{5})(\frac{1}{2})}{\frac{17}{24}}\) To find the value of this fraction, we can first multiply the numerators together and then divide by the denominator: \(P(B|W) = \frac{\frac{1}{10}}{\frac{17}{24}}\) Now we can multiply by the reciprocal of the denominator: \(P(B|W) = \frac{1}{10} \times \frac{24}{17} = \frac{24}{170}\) Finally, we can simplify this fraction: \(P(B|W) = \frac{12}{85}\) So the probability that the coin landed tails (selecting a white ball from urn B) given that a white ball was selected is \(\frac{12}{85}\).

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Most popular questions from this chapter

Suppose that \(E\) and \(F\) are mutually exclusive events of an experiment. Show that if independent trials of this experiment are performed, then \(E\) will occur before \(F\) with probability \(P(E) /[P(E)+\) \(P(F)].\)

Suppose that you continually collect coupons and that there are \(m\) different types. Suppose also that each time a new coupon is obtained, it is a type i coupon with probability \(p_{i}, i=1, \ldots, m .\) Suppose that you have just collected your \(n\)th coupon. What is the probability that it is a new type? Hint: Condition on the type of this coupon.

Consider a sample of size 3 drawn in the following manner: We start with an urn containing 5 white and 7 red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color. Find the probability that the sample will contain exactly. (a) 0 white balls; (b) 1 white ball; (c) 3 white balls; (d) 2 white balls.

In a certain contest, the players are of equal skill and the probability is \(\frac{1}{2}\) that a specified one of the two contestants will be the victor. In a group of \(2^{n}\) players, the players are paired off against each other at random. The \(2^{n-1}\) winners are again paired off randomly, and so on, until a single winner remains. Consider two specified contestants, \(A\) and \(B\), and define the events \(A_{i}, i \leq n, E\) by \(A_{i}:\) \(A\) plays in exactly \(i\) contests: \(E: \quad A\) and \(B\) never play each other. (a) \(\operatorname{Find} P\left(A_{i}\right), i=1, \ldots, n\) (b) Find \(P(E)\) (c) Let \(P_{n}=P(E) .\) Show that $$ P_{n}=\frac{1}{2^{n}-1}+\frac{2^{n}-2}{2^{n}-1}\left(\frac{1}{2}\right)^{2} P_{n-1} $$ and use this formula to check the answer you obtained in part (b). Hint: Find \(P(E)\) by conditioning on which of the events \(A_{i}, i=1, \ldots, n\) occur. In simplifying your answer, use the algebraic identity $$ \sum_{i=1}^{n-1} i x^{i-1}=\frac{1-n x^{n-1}+(n-1) x^{n}}{(1-x)^{2}} $$ For another approach to solving this problem, note that there are a total of \(2^{n}-1\) games played. (d) Explain why \(2^{n}-1\) games are played. Number these games, and let \(B_{i}\) denote the event that \(A\) and \(B\) play each other in game \(i, i=1, \ldots, 2^{n}-1\) (e) What is \(P\left(\bar{B}_{i}\right) ?\) (f) Use part (e) to find \(P(E).\)

Die \(A\) has 4 red and 2 white faces, whereas die \(B\) has 2 red and 4 white faces. A fair coin is flipped once. If it lands on heads, the game continues with dic \(A ;\) if it lands on tails, then dic \(B\) is to be used. (a) Show that the probability of red at any throw is \(\frac{1}{2}.\) (b) If the first two throws result in red, what is the probability of red at the third throw? (c) If red turns up at the first two throws, what is the probability that it is die \(A\) that is being used?
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