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By the third day of a particular week, 2 accidents have already occurred in the intersection. What is the probability that there will be less than a total of 4 accidents during that week

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Answer:

The right answer is "0.70".

Explanation:

The given query seems to be incomplete. Please find below the attachment of the full query.

By using the Bayes' theorem, we get


P[(X<4)|(X \geq 2)] = (P(2 \leq X < 4))/(P(X \geq 2))

By putting the values, we get


=([P(2)+P(3)])/([1-P(0)-P(1)])


=((0.20+0.15))/(1-0.20-0.30)


=(0.35)/(0.5)


=0.70

By the third day of a particular week, 2 accidents have already occurred in the intersection-example-1
User Younes Charfaoui
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