Question

A person wants to construct a hospital in a village for welfare. The probabilities are $0.40$ that some bad elements oppose this work, $0.80$ that the hospital will be completed if there is not any opposition of any bad elements and $0.30$ that the hospital will be completed if bad elements oppose. Determine the probability that the construction of the hospital will be completed.

Answer 1

Hint: First, from the given that a person wants to construct a hospital in a village for welfare.
Some bad elements opposite to this work are the probabilities of $0.40$.
This can be converted in the fraction form as $0.40 = \dfrac{4}{{10}}$
For the not bad elements we have $1 - 0.40$(where one is the overall possibility then minus the bad, we get the not bad oppose)
Hence, we get, not bad oppose is $1 - 0.40 = 0.6 \Rightarrow \dfrac{6}{{10}}$

Complete step by step answer:
From the given, we are able to say the probability of the bad elements opposing is $P({E_1}) = \dfrac{4}{{10}}$where E1 is the bad element. Similarly for the not bad elements oppose we get E2 as, $P({E_2}) = \dfrac{6}{{10}}$
Now take the given hospital construction as A and then the hospital will be complete if there is no opposing of any bad elements is $0.8$, we can take that into the probability, we get $P(\dfrac{A}{{{E_1}}}) = \dfrac{8}{{10}}$where E1 is the bad elements oppose and A is complete in hospital probability.
Similarly for the hospital to be complete in not a bad elements are $0.3$, the probability is $P(\dfrac{A}{{{E_2}}}) = \dfrac{3}{{10}}$
Hence the required is $P(A)$ (probability that when will the construction well get complete)
Therefore, we apply the formula to get the resultant, that is $P(A) = P({E_1}).P(\dfrac{A}{{{E_1}}}) + P({E_2}).P(\dfrac{A}{{{E_2}}})$
Now substituting the all known values we get,
$P(A) = P({E_1}).P(\dfrac{A}{{{E_1}}}) + P({E_2}).P(\dfrac{A}{{{E_2}}}) \Rightarrow \dfrac{4}{{10}}.\dfrac{8}{{10}} + \dfrac{6}{{10}}.\dfrac{3}{{10}}$
Further solving this we get, $P(A) = \dfrac{4}{{10}}.\dfrac{8}{{10}} + \dfrac{6}{{10}}.\dfrac{3}{{10}} \Rightarrow \dfrac{{50}}{{100}} = \dfrac{1}{2}$
Hence $P(A) = \dfrac{1}{2}$ is the probability that the hospital construction will be completed.

Note: Since probability and percentage are interlinked, that the values of the probability written in the decimal mean it is the percentage of the given question that is $P(A) = \dfrac{1}{2} \Rightarrow 0.5$ and then converted into rational form $P(A) = \dfrac{1}{2} \Rightarrow 0.5 = \dfrac{{50}}{{100}}$
Hence the percentage of the given required answer is $50\% $ the complete hospital work contraction.