Question

Three group A, B and C are competing for position on the board of directors of a company. The probabilities of their winning are 0.5, 0.3 and 0.2 respectively. If the group A wins, the probability of introducing a new product in 0.7 and the corresponding probability for group B and C are 0.6 and 0.5 respectively. Find the probability that, new product will be introduced if answer can be expressed as 0.07k then k = ?

Answer 1

Hint: We are given A, B and C. Compete for a position inboard of directors P(A) is 0.5, P(B) is 0.3, and P(C) is 0.2 as A, B, and C are not linked to each other so, they are called the mutually exclusive and exhaustive event.
An event E is dependent on A, B and C for their introduction. So, we use \[P\left( E \right)=P\left( A \right)\cdot P\left( \dfrac{E}{A} \right)+P\left( B \right)\cdot P\left( \dfrac{E}{B} \right)+P\left( C \right)\cdot P\left( \dfrac{E}{C} \right)\] to find the probabilities of the event where $P\left( \dfrac{E}{A} \right)$ is conditional probability given to us already.

Complete step by step answer:
We are given that, there are three groups A, B and C fighting for the position on the Board of Directors. We are given that, probability of group A is chosen as 0.5
\[\Rightarrow P\left( A \right)=0.5\cdots \cdots \cdots \cdots \cdots \left( 1 \right)\]
We have the probability of group B is chosen as 0.3
\[\Rightarrow P\left( B \right)=0.3\cdots \cdots \cdots \cdots \cdots \left( 2 \right)\]
And we have that the probability of group C is chosen as 0.2
\[\Rightarrow P\left( C \right)=0.2\cdots \cdots \cdots \cdots \cdots \left( 3 \right)\]
Now, it is also given that, new product is to be introduced.
Let E be the new product to be introduced.
We are given that, probability E is introduced when A is appointed as 0.7
We can see it clearly, this probability of introducing E is dependent on the condition on which group is chosen. So it is the conditional probability.
Conditional probability is given as $P\left( \dfrac{X}{y} \right)$ where X is event to happen and y is condition it depends on.
So, in our case, our event E is dependent on A, B and C. So, we have
\[\begin{align}
  & P\left( \dfrac{E}{A} \right)=0.7\cdots \cdots \cdots \cdots \cdots \left( 4 \right) \\
 & P\left( \dfrac{E}{B} \right)=0.6\cdots \cdots \cdots \cdots \cdots \left( 5 \right) \\
 & P\left( \dfrac{E}{C} \right)=0.5\cdots \cdots \cdots \cdots \cdots \left( 6 \right) \\
\end{align}\]
Since, selection of A, B or C are total free from the involvement of each other, such event are called mutually exclusive and exhaustive event. So, A, B and C are mutually exclusive and exhaustive event.
So, the total probability at the event to occur is given as
\[P\left( E \right)=P\left( A \right)\cdot P\left( \dfrac{E}{A} \right)+P\left( B \right)\cdot P\left( \dfrac{E}{B} \right)+P\left( C \right)\cdot P\left( \dfrac{E}{C} \right)\]
Now, we use the value of (1), (2), (3), (4), (5), and (6) we get:
\[P\left( E \right)=0.5\times 0.7+0.3\times 0.6+0.2\times 0.5\]
Simplifying we get:
\[P\left( E \right)=0.35+0.17+0.10\]
Solving we get:
\[P\left( E \right)=0.63\]
So the probability that a new event is launched is 0.63.
Now, if the probability is expanded by the number 0.07k then we put $P\left( E \right)=0.07k$ So we get:
\[0.07k=0.63\]
Divide both sides by 0.07 we get:
\[\dfrac{0.07k}{0.07}=\dfrac{0.63}{0.07}\]
So k = 9, so we get value of k as 9.

Note:
 We have to be clear while reading that, as we are not given that there is any relation between A, B or C. So, it means they are not connected. So, they are mutually exclusive.
Also, while solving the probabilities, remember that, when we use the product, two or more decimal numbers the final number will have a decimal term as the sum of all the decimal term in total. Like,
\[0.7\times 0.5=0.35\]
In $0.7\times 0.5$ there are 2 decimals, one at 0.7 and one at 0.5 so, the product has decimal after 2 terms.