It is reported that dog bites are somewhat common among tourists who visit Thailand due to the stray dog population. One study showed that 13 out of 1000 tourists get bit by a dog. On average, the percentage of tourists who have been bit by a dog that get rabies is around 15%. Ninety-nine percent of tourists who do not get bit by a dog will not get rabies.

1. Define the two events in this problem.

A=
B=

2. Using proper probability notation, identify the following values:

a. 0.013 =
b. 0.15 =
c. 0.99 =

3. Create and upload an image of a hypothetical two-way table for this scenario.
4. Create and upload an image of a tree diagram for this scenario.
5. Using either your table or tree, calculate the probability that a randomly selected tourist will end up getting rabies. Upload a picture of your work and place your final answer (rounded to three decimal places) in the blank provided. Include proper probability notation in the work that you upload.
6. Using either your table or tree, calculate the probability that a tourist who does not get rabies then gets bit by a dog. Upload a picture of your work and place your final answer (rounded to three decimal places) in the blank provided. Include proper probability notation in the work that you upload.

"The percentage of tourists who have been bitten by a dog that get rabies is 15%" i.e. Given that the tourist was bitten by a dog, he got rabies. This is a conditional probability, is the probability of "B" given that "A" has already happened:

P(B|A)= 0.15

c.

"Ninety-nine percent of tourists who do not get bit by a dog will not get rabies."

The event "The tourist did not get bitten by a dog" is complementary to "A", so I'll symbolize it as: [tex]A^c[/tex]

The event "The tourist did not get rabies" is complementary to the event "B", so I'll symbolize it as [tex]B^c[/tex]

[tex]P(B^c|A^c)= 0.99[/tex]

3) See attachment for table

P(A)= 0.013 ⇒ P([tex]A^c[/tex])= 1 - 0.013= 0.987

P(A∩B)= P(B|A)*P(A)= 0.15*0.013= 0.00195 = 0.002

P(A∩[tex]B^c[/tex])= P(A) - P(A∩B)= 0.013-0.00195= 0.01105 = 0.011

[tex]P(B^c|A^c)= 0.99[/tex]

P([tex]B^c[/tex]∩[tex]A^c[/tex])= P([tex]A^c[/tex]) * [tex]P(B^c|A^c)[/tex]= 0.987*0.99= 0.977

P([tex]B^c[/tex])= P(A∩[tex]B^c[/tex]) + P([tex]B^c[/tex]∩[tex]A^c[/tex])= 0.011 + 0.977= 0.988

P(B∩[tex]A^c[/tex])= P([tex]A^c[/tex]) - P([tex]B^c[/tex]∩[tex]A^c[/tex])= 0.987 - 0.977= 0.01

5) P(B)= P(A∩B) + P(B∩[tex]A^c[/tex])= 0.002 + 0.01= 0.012

4) Attch

6)

P(A|[tex]B^c[/tex])= P(A∩[tex]B^c[/tex]) = 0.011 = 0.0111

P([tex]B^c[/tex]) 0.988

I hope this helps!