Question

This is for my Probability and Statistic class,

I need help with this probability problem. We just vaguely just learned conditional and unconditional rules. But I still don't get it.

This is the question and sub-questions involved in the problem. (Given labeled in-class data tables).

Conditional probabilities:

a) Determine the conditional probability that A has visited the (randomly selected) state, given that B has visited it. Is this conditional probability greater than, less than, or the same as the ( unconditional) probability that A has visited the state?

b) Determine the conditional probability that B has visited the (randomly selected) state, given that A has visited it. Is this conditional probability greater than, less than, or the same as the ( unconditional) probability that B has visited the state?

c) Are the events {A has visited the state}, and {B has visited the state} mutually exclusive? Justify your answer.

d) Are the events {A has visited the state}, and {B has visited the state} independent? Justify your answer.

e) Determine the conditional probability that both persons have visited the (randomly selected) state, given that at least one of them has visited it.

Thank you,

Person A and Person B should look through the list and check off which states you have ever visited. (Let's define "visited" to mean that you have ever set foot in the state.) 1) Produce a table like the following that lists counts for the number of states visited by Person A and Person B: Person B has visited Person B has not visited | Total Person A has visited 5 Person A has not visited 42 45 Total 50 Now suppose that one of the 50 states is selected at random, so each state has a 1/50 = .02 probability of being selected. (50 states in the United States)

Answer 1

Let as first define the conditional probability of evente AA and B. - P(A/B) ces ginien the event B, what is the probability of occusance of event A. For two events A and B to be mutually exclusive, P(A and B) : , P (ANB)=0. or P(AIB)- PCA). P(AIB)=P(AMB) (By definitions) PCB) Two events are independent of P(ANB)= PCAPCB): From the table we can write the probabilities, PCA visted) = 8/50 = %10 P(B visited) - 5/0 = Yo I Similarly sted) methodno ---PCA visited) -1-7700- 50,0 P(B not vistited )- %0 (a) conditional probability that A visted, given that B visited = pl = P(AIB)- Pchaude) P(B)

from the table , P(A visted and Bristed2. S P(ARB) = 245 - P(A/B)= P(AMB) _ %0 2 PCBD 5% - % to unconditional probatrility P(A)=5750 = 40 PLASB)= 2% > P(A)=%o ce, conditional - greater than encondition b Similarly, conditional probability that B risted ginen A visited I Phan = P(B(A)= PCB and a visted) PLA visted) PCA) T = 7 = P(ANB) = %0 Now unconditional poobabi Wtry of B visited PCB)<%%

Now conditional probability, PROVE PIBIA) = 45 > PCB)-ho e., conditional probability is greater than unconditional probability . c) A and B are mutually exclusive if P(A and B) = P (AMB) =0. here PlanB)=2/ (from the table) . ... A and B are not mutually exclensine (6) of A and B are independent events if PLANB) - PCA) PCB) Here PLANB)=2/50 PCA) PCB) = 560*% - 3* - I #3 $*10*5*10 100 » A and B are not independent. 16 Now we have to find the probability that both visited given that atleast one visited e-, Pl A and B / foo Boo Both - RCADA ac Raroto BABY AB)

PLAB|Aor or AB) = P(ABIA TO ABIB or ABI AB) = P(ABlA ) +P(ABIB) + PCABIAB) = P(ABNA), PLABOB), P (AB) - PLA) PIB) = P(AB) + PCAB) + P(AB) PCA PCBD = 25+ %5 + %0 = 16 = 4042. 42 SO - 12 50 50 = 21