Answer to Question #330801 in Statistics and Probability for BERNARD

Question #330801

n a farming community, 30% of the farmers grow oranges only, 10% grow lemons only and 4% grow both oranges and lemons. 1.1) What proportion of farmers in the community grow either oranges or lemons? [3 marks] 1.2) If a farmer is chosen randomly from these in the community, what is the probability that he grows neither oranges nor lemons? [marks] 1.3) Of all the farmers who grow oranges, what proportion grow lemons also?


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Expert's answer
2022-04-21T15:14:23-0400

Let AA- the event that a farmer grows oranges only, P(A)=0.3;P(A) =0.3;

BB- the event that a farmer grows lemons only, P(B)=0.1;P(B) =0.1;

CC- the event that a farmer grows both oranges and lemons, P(C)=0.04.P(C) =0.04.


1.1) DD- the event that a farmer grows either oranges or lemons, D=AB.D=A\cup B.

Events A,BA, B are mutually exclusive events,

P(D)=P(AB)=P(A)+P(B)==0.30+0.10=0.40=40%.P(D) =P(A\cup B) =P(A) +P(B) =\\ =0.30+0.10=0.40=40\%.


1.2) EE- the event that a farmer grows neither oranges nor lemons.

FF- the event that a farmer grows oranges or lemons or both, F=ABC;F=A\cup B\cup C;

Events A,B,CA, B, C are mutually exclusive events,

P(F)=P(ABC)==P(A)+P(B)+P(C)==0.30+0.10+0.04=0.44=44%.P(F) =P(A\cup B\cup C) =\\=P(A) +P(B) +P(C) =\\ =0.30+0.10+0.04=0.44=44\%.


Events EE and FF are complementary events,

E=Fˉ,P(E)=1P(F)=10.44=0.56=56%.E=\bar F, P(E) =1-P(F)=1-0.44=0.56=56\%.


1.3)

P(C)P(A)+P(C)=0.040.3+0.04=0.1176=11.76%.\cfrac{P(C) } {P(A)+P(C) } =\cfrac{0.04} {0.3+0.04} =0.1176=11.76\%.


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