Speeding linked to fatal car crashes. According to the National Highway Traffic and Safety Administration’s National Center for Statistics and Analysis (NCSA), “Speeding is one of the most prevalent factors contributing to fatal traffic crashes” (NHTSA Technical Report, August 2005). The probability that speeding is a cause of a fatal crash is .3. Furthermore, the probability that speeding and missing a curve are causes of a fatal crash is .12. Given speeding is a cause of a fatal crash, what is the probability that the crash occurred on a curve?

Short Answer

Expert verified

The probability is 0.4.

Step by step solution

01

Important formula

The formula is P(AB)=P(A).P(B|A).

02

The probability that the crash occurred on a curve.

Here, P(G)=0.3=Speeding is a cause of a fatal crash.

P(F|G)=0.12=Speeding and missing a curve are causes of a fatal crash.

P(GF)=P(G)×P(F|G)0.12=(0.38)P(F|G)P(F|G)=0.4

Therefore the probability is 0.4.

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Most popular questions from this chapter

Suppose the events B1and B2are mutually exclusive and complementary events, such thatP(B1)=.75andP(B2)=.25 Consider another event A such that role="math" localid="1658212959871" P(AB1)=.3, role="math" localid="1658213029408" P(AB2)=.5.

  1. FindP(B1A).
  2. FindP(B2A)
  3. Find P(A) using part a and b.
  4. Findrole="math" localid="1658213127512" P(B1A).
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b. If the gas station air pressure gauge reads 55 psi, what is the probability that the pressure is over-reported by 8 psi or more?

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d. Are the events A= {over report by 4 psi or more} and B= {over report by 6 psi or more} mutually exclusive?

e.Based on your answer to part d, why do the probabilities in the table not sum to 1?

In a random sample of 106 social (or service) robots designed to entertain, educate, and care for human users, 63 were built with legs only, 20 with wheels only, 8 with both legs and wheels, and 15 with neither legs nor wheels. One of the 106 social robots is randomly selected and the design (e.g., wheels only) is noted.

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The outcomes of two variables are (Low, Medium, High) and (On, Off), respectively. An experiment is conducted in which the outcomes of each of the two variables are observed. The accompanying two-way table gives the probabilities associated with each of the six possible outcome pairs.

Low

Medium

High

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.50

.10

.05

Off

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Consider the following events:

A: {On}

B: {Medium or on}

C: {Off and Low}

D: {High}

a. Find P (A).

b. Find P (B).

c. Find P (C).

d. Find P (D).

e. FindP(AC).

f. FindP(AB).

g. FindP(AB).

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b.Find the probability that EMS can meet the demand for an ambulance at location B.

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