# a) Mr Okyere takes banana and apple to the office to share with his of

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.
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 i $$\displaystyle{\frac{{{1}}}{{{5}}}},\ {\frac{{{1}}}{{{3}}}}$$ and $$\displaystyle{\frac{{{7}}}{{{15}}}}$$ A farmer is randomly selected from the group of farmers. What is the probability that the farmer chose none of the chemicals?
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.

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Michele Grimsley
Step 1
Note:
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Step 2
Concept of conditional probability:
For any two events A and B with $$\displaystyle{P}{\left({B}\right)}{>}{0}$$, the formula for conditional probability is given by:
$$\displaystyle{P}{\left({A}{\mid}{B}\right)}={\frac{{{P}{\left({A}\cap{B}\right)}}}{{{P}{\left({B}\right)}}}}$$
Step 3
From the given information, the probability that the person will eat the banana is 0.7. That is, $$\displaystyle{P}{\left({A}\right)}={0.7}$$.
The probability that the person will eat the apple is 0.8. That is, $$\displaystyle{P}{\left({B}\right)}={0.8}$$.
The probability that the person will eat banana and apple is 0.6. That is, $$\displaystyle{P}{\left({A}{\quad\text{and}\quad}{B}\right)}={0.6}$$.
It is required to find the probability that he will eat the apple given that he does not eat the banana is $$\displaystyle{P}{\left({B}{\mid}{A}^{{{C}}}\right)}$$ and is calculated as follows:
$$\displaystyle{P}{\left({B}{\mid}{A}^{{{C}}}\right)}={\frac{{{P}{\left({B}\cap{A}^{{{C}}}\right)}}}{{{P}{\left({A}^{{{C}}}\right)}}}}$$
$$\displaystyle={\frac{{{P}{\left({B}\right)}-{P}{\left({A}\cap{B}\right)}}}{{{1}-{P}{\left({A}\right)}}}}$$
$$\displaystyle={\frac{{{0.8}-{0.6}}}{{{1}-{0.7}}}}$$
$$\displaystyle={\frac{{{0.2}}}{{{0.3}}}}$$
$$\displaystyle={0.6667}$$
Therefore, the conditional probability that he will eat the apple given that he does not eat the banana is 0.6667.