Answer to Question #331104 in Statistics and Probability for angie 123

Question

In 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? [2 marks]

1.3) Of all the farmers who grow oranges, what proportion grow lemons also?  [2 marks]

Answer 1

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

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

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

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

Events $A, B$ are mutually exclusive events,

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

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

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

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

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

Events $E$ and $F$ are complementary events,

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

1.3)

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