class 12 maths probability

An item is manufactured by three machines $A,{\rm{ }}B and C$. Out of the total number of items manufactured during a specified period, $50\%$ are manufactured on $A,$ $30\%$ on $B$ and $20\%$ on $C$. $2\%$ of the items produced on $A$ and $2\%$ of items produced on $B$ are defective and $3\%$ of these produced on $C$ are defective. All the items are stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine $A$ ?

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📘 Probability NCERT,Exemp,Q.48,Page.277 LA

An item is manufactured by three machines $A,{\rm{ }}B and C$. Out of the total number of items manufactured during a specified period, $50\%$ are manufactured on $A,$ $30\%$ on $B$ and $20\%$ on $C$. $2\%$ of the items produced on $A$ and $2\%$ of items produced on $B$ are defective and $3\%$ of these produced on $C$ are defective. All the items are stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine $A$ ?

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

Let ${E_1} =$ Event that item is manufactured on $A$,

${E_2} =$ Event that an item is manufactured on $B$,

${E_3} =$ Event that an item is manufactured on $C$,

Let $E$ be the event that an item is defective.

$\therefore$ $\quad P\left( {{E_1}} \right) = \frac{{50}}{{100}} = \frac{1}{2},P\left( {{E_2}} \right) = \frac{{30}}{{100}} = \frac{3}{{10}}$ and $P\left( {{E_3}} \right) = \frac{{20}}{{100}} = \frac{1}{5}$

$P\left( {\frac{E}{{{E_1}}}} \right) = \frac{2}{{100}} = \frac{1}{{50}},P\left( {\frac{E}{{{E_2}}}} \right) = \frac{2}{{100}} = \frac{1}{{50}}$ and $P\left( {\frac{E}{{{E_3}}}} \right) = \frac{3}{{100}}$

$\therefore$ $P\left( {\frac{{{E_1}}}{E}} \right) = \frac{{P\left( {{E_1}} \right) \cdot P\left( {\frac{E}{{{E_1}}}} \right)}}{{P\left( {{E_1}} \right) \cdot P\left( {\frac{E}{{{E_1}}}} \right) + P\left( {{E_2}} \right) \cdot P\left( {\frac{E}{{{E_2}}}} \right) + P\left( {{E_3}} \right) \cdot P\left( {\frac{E}{{{E_3}}}} \right)}}$

$= \frac{{\frac{1}{2} \cdot \frac{1}{{50}}}}{{\frac{1}{2} \cdot \frac{1}{{50}} + \frac{3}{{10}} \cdot \frac{1}{{50}} + \frac{1}{5} \cdot \frac{3}{{100}}}}$

$= \frac{{\frac{1}{{100}}}}{{\frac{1}{{100}} + \frac{3}{{500}} + \frac{3}{{500}}}} = \frac{{\frac{1}{{100}}}}{{\frac{{5 + 3 + 3}}{{500}}}} = \frac{5}{{11}}$

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