Probability question

Question

: (a) Mr Okyere takes banana and apple to the office to share with his office

mate. The probability that he will eat the banana is 0.7 and 0.8 is the probability

that he will eat the apple. The probability that he will eat both fruits is 0.6. Find

the conditional probability that he will eat the apple given that he does not eat the

banana. [3 Marks]

(b) The Ministry of Agriculture in its fight against the army worm infection of maize

farms in the country has approved three brands of chemicals A, B and C to be used

by farmers as follows: Farmers are to combine exactly two of the chemicals to spray

their farms or they should use none of them. The proportions of farmers who opted

for chemicals A, B and C are respectively given by 1/5,

1/3 and 7/15 . A farmer is randomly

selected from the group of farmers. What is the probability that the farmer chose

none of the chemicals? [8 Marks]

(c) Consider three boxes containing a brand of light bulbs. Box I contains 6 bulbs

of which 2 are defective, Box 2 has 1 defective and 3 functional bulbs and Box 3

contains 3 defective and 4 functional bulbs. A box is selected at random and a bulb

drawn from it at random is found to be defective. Find the probability that the box

selected was Box 2.

Answer 1

a)

Let P(eat apple) = A and P(eat banana) = B

P(A) = .8

P(B) = .7

P(AB) = .6

P(A|B') = P(AB')/P(B')

P(B') = 1 - P(B) = 1 - .7 = .3

P(AB') = P(A) - P(AB) = .8 - .6 = .2

P(A|B') = .2/.3 = 2/3

b)P(farmer chose none of the chemicals) = 1 - P(farmer chose at least one of the chemicals)

E(chemicals selected) = P(farmer selects a chemical)*E(chemicals selected|farmer selects a chemical)

E(chemicals selected) = 1/5 + 1/3 + 7/15 = 1

E(chemicals selected|farmer selects a chemical) = 2 (farmers have to select 2 chemicals if they select any)

Then, P(farmer selects a chemical) = E(chemicals selected)/E(chemicals selected|farmer selects a chemical) = (1)/(2) = 1/2

Then, P(farmer does not select a chemical) = 1 - 1/2 = 1/2

There is another way that you can work this out that you may prefer.

P(C) = P(AB'C) + P(A'BC) = 7/15 , as P(ABC) = P(A'B'C) = 0 - one and three chemicals are never selected.

P(B) = P(ABC') + P(A'BC) = 1/3

P(A) = P(ABC') + P(AB'C) = 1/5

Add the three lines, and

2(P(AB'C) + P(A'BC)+P(ABC')) = 7/15+1/3+1/5 = 1, so

P(AB'C) + P(A'BC)+P(ABC') = 1/2

P(ABC) = P(AB'C') = P(A'BC') = P(A'B'C) = 0 as 2 chemicals are always bought.

P(A U B U C) = P(AB'C) + P(A'BC)+P(ABC') +P(ABC) + P(AB'C') + P(A'BC') + P(A'B'C) =

(P(AB'C) + P(A'BC)+P(ABC') ) + P(ABC) + P(AB'C') + P(A'BC') + P(A'B'C) =

1/2 + 0 + 0 + 0 + 0 = 1/2

P(A'B'C') = 1 - P(A U B U C) = 1 - 1/2 = 1/2

c) P(defective|Box 1) = 2/6 = 1/3

P(defective|Box 2) = 1/4

P(defective|Box 3) = 3/7

P(Box 1) = P(Box 2) = P(Box 3) = 1/3 A box is selected at random. Without any more information, one supposes this to mean that one is equally likely to select each box.

Then, P(Box 2|defective) = P(Box 2 and defective)/P(defective) =

P(defective|Box 2)* P(Box 2)/(P(defective|Box 1)* P(Box 1)+P(defective|Box 2)* P(Box 2)+P(defective|Box 3)* P(Box 3)) = 1/3*1/4/(1/3*1/3+1/4*1/3+3/7*1/3) = 1/3*1/4/(1/3 *(1/3+1/4+3/7)) =

1/4/(1/3+1/4+3/7) = 21/84/(28/84+21/84+36/84) = 21/(28+21+36) = 21/85