Question 1196656


given:

Stage 1 has two nodes and stage 2 has four nodes.
In stage 1, the branch from the starting position to node {{{A}}} is labeled {{{0.3}}}. The branch from the starting position to node {{{B}}} is an answer {{{blank}}}.

In stage 2, the branch from node {{{A}}} to node {{{C}}} is an answer {{{blank}}}. The branch from node {{{A}}} to node{{{ D}}} is labeled{{{ 0.4}}}.

In stage 2, the branch from node {{{B}}} to node {{{C}}} is labeled{{{ 0.2}}}. The branch from node {{{B }}}to node{{{ D }}}is an answer {{{blank}}}.


recall: The sum of the probabilities on the branches leaving any node is {{{always}}} {{{1}}}. This allows us to fill in the middle two blanks.


then
In stage 1, the branch from the starting position to node {{{A}}} is labeled {{{0.3}}}. The branch from the starting position to node {{{B }}} will be {{{0.7}}}.

In stage 2, the branch from node {{{A}}} to node{{{ C}}} is{{{ 0.6}}}. The branch from node {{{A}}} to node {{{D}}} will be labeled {{{0.4}}}.

In stage 2, the branch from node {{{B }}}to node {{{C}}} is labeled {{{0.2}}}. The branch from node {{{B}}} to node {{{D}}} will be{{{ 0.8}}}.



Outcome

P(A ∩ C) ={{{0.3*0.6=0.18}}}
P(A ∩ D) ={{{0.3*0.4=0.12}}}
P(B ∩ C) ={{{0.7*0.2=0.14}}}
P(B ∩ D) ={{{0.7*0.8=0.56}}}