Probabilities?
During a safari in Tanzania, three animals can be spotted: elephants, wildebeest, and lions. Safari operators have long-standing experience that 17% of the time will see a wildebeest and an elephant, 26% will see a lion and a wildebeest, and 20% will see a lion and an elephant. The only chance of seeing an elephant is 13%. Let E be the event of seeing an elephant, G the event of seeing a wildebeest, and L the event of seeing a lion. P(G) = 53% and P(E) = 45%.
(a) Calculate P(EUG). b) Determine P(G|E) and P(E|G). c) Calculate P(E∩L∩G).
So far I have the following ideas:
Regarding the last one, I have doubts as to whether the solution, let alone the calculation method, is correct.
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In the case of the conditional probabilities, you set the equations correctly, but in the denominator incorrect values! P(E) is 0.45, so P(G|E)=0.17/0.45=17/45≅37.78 %
and P(E|G)=0.17/0.53≅32.08 %
Generally, I would create a tree chart here, e.g. beginning with the branches for E, then G, then L.
The probabilities of the branches E (=0.45) and E-Strich (=0.55) are predetermined. The Ast E->G is P(G|E), i.e. the aforementioned 17/45. E->G-Strich 1-17/45=28/45 is accordingly.
Only E, i.e. P(E n G-Strich n L-Strich) is specified with 0.13, i.e. You can now easily calculate the associated Ast L-Strich, and then the above-lying Ast L.
P(G)=P(E n G) + P(E-Strich n G)=0.53: you can calculate the Ast P(G|E-Strich).
With the other specifications, you are in a similar way. So lion and elephant equal to 20% means that the paths (E,G,L) and (E,G-Strich,L) must yield 0.2. You already have the last path after determining the branches using “only E”, i.e. You only lack the Ast L of the Path (E,G,L). And you have calculated this branch, then you automatically have the solution to P(E n G n L), that is the path (E,G,L). (according to my bill, 5% would have to come out)