Answer

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**Hint:**In this question we will consider the cases of the probabilities as $4$ separate values and then apply Bayes theorem which tells us the conditional probability of an event given that one event has already occurred. In this question we have to find what the probability is for a television to be from plant $2$ given that the television is already found to be of standard quality.

**Complete step by step answer:**

We know that plant $1$ manufactures $70\%$ of the total televisions in the company. Consider ${{E}_{1}}$ to be all the televisions manufactured by plant $1$, it can be written as:

$P({{E}_{1}})=\dfrac{70}{100}$

We know that plant $2$ manufactures $30\%$ of the total televisions in the company. Consider ${{E}_{2}}$ to be all the televisions manufactured by plant $2$, it can be written as:

$P({{E}_{1}})=\dfrac{30}{100}$

Now we know that that from plant $1$, $80\%$ of the televisions are of standard quality therefore, we can write it as:

$P\left( \dfrac{A}{{{E}_{1}}} \right)=\dfrac{80}{100}$

Now we know that that from plant $2$, $90\%$ of the televisions are of standard quality therefore, we can write it as:

$P\left( \dfrac{A}{{{E}_{2}}} \right)=\dfrac{90}{100}$

Now we have to find the probability for a television to be from $2$ given that the television is already found to be of standard quality.

This can be found out by multiplying the probabilities of the television being from the plant $2$and the probability of the television to be of standard quality from plant $2$, which should be divided by total sample space which is product of probabilities of the television being from the plant $2$and the probability of the television to be of standard quality from plant $2$ added with probabilities of the television being from the plant $1$and the probability of the television to be of standard quality from plant $1$.

Mathematically, it can be written as:

$P\left( \dfrac{{{E}_{2}}}{A} \right)=\dfrac{P\left( {{E}_{2}} \right)P\left( \dfrac{A}{{{E}_{2}}} \right)}{P\left( {{E}_{1}} \right)P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)P\left( \dfrac{A}{{{E}_{2}}} \right)}$

On substituting the values, we get:

$P\left( \dfrac{{{E}_{2}}}{A} \right)=\dfrac{\dfrac{30}{100}\times \dfrac{90}{100}}{\dfrac{70}{100}\times \dfrac{80}{100}+\dfrac{30}{100}\times \dfrac{90}{100}}$

On multiplying all the terms and taking a common denominator, we get:

$P\left( \dfrac{{{E}_{2}}}{A} \right)=\dfrac{\dfrac{2700}{100}}{\dfrac{5600+2700}{100}}$

Now on cancelling the denominators, we get:

$P\left( \dfrac{{{E}_{2}}}{A} \right)=\dfrac{2700}{5600+2700}$

On adding the terms in the denominator, we get:

$P\left( \dfrac{{{E}_{2}}}{A} \right)=\dfrac{2700}{8300}$

On simplifying, we get:

$P\left( \dfrac{{{E}_{2}}}{A} \right)=\dfrac{27}{83}$, which is the required solution.

**Note:**The general formula of bayes theorem should be remembered which is $P(A/B)=\dfrac{P(A)P(B/A)}{P(B)}$ which indicates the probability of the event $A$ occurring given that event $B$ occurred. If the occurrence of $B$ does not affect the probability of $A$, then $A$ and $B$ are mutually exclusive events.

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