3.1 Problems on conditional probability (Page 3/3)

Applied probability Page 3 / 3

A quality control group is designing an automatic test procedure for compact disk players coming from a production line. Experience shows thatone percent of the units produced are defective. The automatic test procedure has probability 0.05 of giving a false positive indication and probability0.02 of giving a false negative. That is, if D is the event a unit tested is defective, and T is the event that it tests satisfactory, then $P (T | D) = 0.05$ and $P (T^{c} | D^{c}) = 0.02$ . Determine the probability $P (D^{c} | T)$ that a unit which tests good is, in fact, free of defects.

\frac{P (D^{c} | T)}{P (D | T)} = \frac{P (T | D^{c}) P (D^{c})}{P (T | D) P (D)} = \frac{0.98 \cdot 0.99}{0.05 \cdot 0.01} = \frac{9702}{5}

P (D^{c} | T) = \frac{9702}{9707} = 1 - \frac{5}{9707}

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Five boxes of random access memory chips have 100 units per box. They have respectively one, two, three, four, and five defective units. A box is selected at random,on an equally likely basis, and a unit is selected at random therefrom. It is defective. What are the (conditional) probabilities the unit was selected from each of the boxes?

$H_{i} =$ the event from box i . $P (H_{i}) = 1 / 5$ and $P (D | H_{i}) = i / 100$ .

P (H_{i} | D) = \frac{P (D | H_{i}) P (H_{i})}{\sum P (D | H_{j}) P (H_{j})} = i / 15, 1 \leq i \leq 5

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Two percent of the units received at a warehouse are defective. A nondestructive test procedure gives two percent false positive indicationsand five percent false negative. Units which fail to pass the inspection are sold to a salvage firm. This firm applies a corrective procedurewhich does not affect any good unit and which corrects 90 percent of the defective units. A customer buys a unit from the salvage firm. Itis good. What is the (conditional) probability the unit was originally defective?

Let $T =$ event test indicates defective, $D =$ event initially defective, and $G =$ event unit purchased is good. Data are

P (D) = 0.02, P (T^{c} | D) = 0.02, P (T | D^{c}) = 0.05, P (G T^{c}) = 0,

P (G | D T) = 0.90, P (G | D^{c} T) = 1

P (D | G) = \frac{P (G D)}{P (G)}, P (G D) = P (G T D) = P (D) P (T | D) P (G | T D)

P (G) = P (G T) = P (G D T) + P (G D^{c} T) = P (D) P (T | D) P (G | T D) + P (D^{c}) P (T | D^{c}) P (G | T D^{c})

P (D | G) = \frac{0.02 \cdot 0.98 \cdot 0.90}{0.02 \cdot 0.98 \cdot 0.90 + 0.98 \cdot 0.05 \cdot 1.00} = \frac{441}{1666}

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At a certain stage in a trial, the judge feels the odds are two to one the defendent is guilty. It is determined that the defendent is left handed. An investigatorconvinces the judge this is six times more likely if the defendent is guilty than if he were not. What is the likelihood, given this evidence, that the defendent is guilty?

Let $G =$ event the defendent is guilty, $L =$ the event the defendent is left handed. Prior odds: $P (G) / P (G^{c}) = 2$ . Result of testimony: $P (L | G) / P (L | G^{c}) = 6$ .

\frac{P (G | L)}{P (G^{c} | L)} = \frac{P (G)}{P (G^{c})} \cdot \frac{P (L | G)}{P (L | G^{c})} = 2 \cdot 6 = 12

P (G | L) = 12 / 13

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Show that if $P (A | C) > P (B | C)$ and $P (A | C^{c}) > P (B | C^{c})$ , then $P (A) > P (B)$ . Is the converse true? Prove or give a counterexample.

$P (A) = P (A | C) P (C) + P (A | C^{c}) P (C^{c}) > P (B | C) P (C) + P (B | C^{c}) P (C^{c}) = P (B)$ .

The converse is not true. Consider $P (C) = P (C^{c}) = 0.5$ , $P (A | C) = 1 / 4$ ,

$P (A | C^{c}) = 3 / 4$ , $P (B | C) = 1 / 2$ , and $P (B | C^{c}) = 1 / 4$ . Then

1 / 2 = P (A) = \frac{1}{2} (1 / 4 + 3 / 4) > \frac{1}{2} (1 / 2 + 1 / 4) = P (B) = 3 / 8

But $P (A | C) < P (B | C)$ .

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Since $P (\cdot | B)$ is a probability measure for a given B , we must have $P (A | B) + P (A^{c} | B) = 1$ . Construct an example to show that in general $P (A | B) + P (A | B^{c}) \neq 1$ .

Suppose $A \subset B$ with $P (A) < P (B)$ . Then $P (A | B) = P (A) / P (B) < 1$ and $P (A | B^{c}) = 0$ so the sum is less than one.

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Use property (CP4) to show

$P (A | B) > P (A) iff P (A | B^{c}) < P (A)$
$P (A^{c} | B) > P (A^{c}) iff P (A | B) < P (A)$
$P (A | B) > P (A) iff P (A^{c} | B^{c}) > P (A^{c})$

$P (A | B) > P (A)$ iff $P (A B) > P (A) P (B)$ iff $P (A B^{c}) < P (A) P (B^{c})$ iff $P (A | B^{c}) < P (A)$
$P (A^{c} | B) > P (A^{c})$ iff $P (A^{c} B) > P (A^{c}) P (B)$ iff $P (A B) < P (A) P (B)$ iff $P (A | B) < P (A)$
$P (A | B) > P (A)$ iff $P (A B) > P (A) P (B)$ iff $P (A^{c} B^{c}) > P (A^{c}) P (B^{c})$ iff $P (A^{c} | B^{c}) > P (A^{c})$

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Show that $P (A | B) \geq (P (A) + P (B) - 1) / P (B)$ .

$1 \geq P (A \cup B) = P (A) + P (B) - P (A B) = P (A) + P (B) - P (A | B) P (B)$ . Simple algebra gives the desired result.

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Show that $P (A | B) = P (A | B C) P (C | B) + P (A | B C^{c}) P (C^{c} | B)$ .

P (A | B) = \frac{P (A B)}{P (B)} = \frac{P (A B C) + P (A B C^{c})}{P (B)}

= \frac{P (A | B C) P (B C) + P (A | B C^{c}) P (B C^{c})}{P (B)} = P (A | B C) P (C | B) + P (A | B C^{c}) P (C^{c} | B)

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An individual is to select from among n alternatives in an attempt to obtain a particular one. This might be selection from answers on amultiple choice question, when only one is correct. Let A be the event he makes a correct selection, and B be the event he knows which is correct before making the selection. We suppose $P (B) = p$ and $P (A | B^{c}) = 1 / n$ . Determine $P (B | A)$ ; show that $P (B | A) \geq P (B)$ and $P (B | A)$ increases with n for fixed p .

$P (A | B) = 1$ , $P (A | B^{c}) = 1 / n$ , $P (B) = p$

P (B | A) = \frac{P (A | B) P (B)}{P A | B) P (B) + P (A | B^{c}) P (B^{c})} = \frac{p}{p + \frac{1}{n} (1 - p)} = \frac{n p}{(n - 1) p + 1}

\frac{P (B | A)}{P (B)} = \frac{n}{n p + 1 - p} increases from 1 to 1 / p as n \to \infty

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Polya's urn scheme for a contagious disease . An urn contains initially b black balls and r red balls ( $r + b = n$ ). A ball is drawn on an equally likely basis from among those in the urn, thenreplaced along with c additional balls of the same color. The process is repeated. There are n balls on the first choice, $n + c$ balls on the second choice, etc. Let B _k be the event of a black ball on the k th draw and R _k be the event of a red ball on the k th draw. Determine

$P (B_{2} | R_{1})$
$P (B_{1} B_{2})$
$P (R_{2})$
$P (B_{1} | R_{2})$ .

$P (B_{2} | R_{1}) = \frac{b}{n + c}$
$P (B_{1} B_{2}) = P (B_{1}) P (B_{2} | B_{1}) = \frac{b}{n} \cdot \frac{b + c}{n + c}$
$P (R_{2}) = P (R_{2} | R_{1}) P (R_{1}) + P (R_{2} | B_{1}) P (B_{1})$
$= \frac{r + c}{n + c} \cdot \frac{r}{n} + \frac{r}{n + c} \cdot \frac{b}{n} = \frac{r (r + c + b)}{n (n + c)}$
$P (B_{1} | R_{2}) = \frac{P (R_{2} | B_{1}) P (B_{1})}{P (R_{2})}$ with $P (R_{2} | B_{1}) P (B_{1}) = \frac{r}{n + c} \cdot \frac{b}{n}$ . Using (c), we have
$P (B_{1} | R_{2}) = \frac{b}{r + b + c} = \frac{b}{n + c}$

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Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

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