1. (60 pts) According to distraction-gov, 10% of all fatal crashes involve drive

Question

1. (60 pts) According to distraction-gov, 10% of all fatal crashes involve drivers who were distracted (for example, using a cell phone). In other words, if you are in a fatal crash, then there is a 10% chance you were distracted. The percentage of drivers who are distracted is estimated to be 5% (a) Using Bayes' Rule, what is the formula for P(F|D)? Gust the notation, don't try to plug numbers in yet!) (b) How much does driving distracted increase your probability of being in a fatal car crash? Hint: The question is asking for the ratio: P(FD) PE D ) = increase in chance of a fatal crash due to using a cell phone First, show the expansion of the numerator and denominator using Bayes' rule, then calculate. Notice, you don't know P(F), but you don't need it. Your answer should now be a number, but remember, you don't have to simplify it

Explanation / Answer

A) P(F | D) = P(D | F) * P(F) / P(D)

B) P(F | D) / P(F | Dc) = (P(D | F) * P(F) / P(D) ) / (P (Dc | F) * P (F ) / P (Dc))

= P(D | F ) * P(Dc) / P (D) * P ( Dc | F)

= 0.1 * 0.95 / 1 - (0.1 * 0.95) = 0.095 / 0.905 = 0.105

C)If P(D | F) = P(D), then D and F will be independent.

P(D | F) = 0.1

P(D) = 0.05

As they are not equal, so D and F are not independent.

  

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