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# In a factory which manufactures bolts, machines A, B and C manufacture $30%,50%$ and $20%$ of the bolts respectively. Of their output $3%,4%$ and $1%$ respectively are defective bolts. A bolt is drawn at random from the product and is found to be defective. Find the probability that this is not manufactured by machine B. Verified
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Hint: Calculate the probability of each of the events of manufacturing a bolt from the three machines and manufacturing a default bolt. Use conditional probability to calculate the probability of getting a defective bolt. Use the fact that the sum of probability of occurrence of an event and non occurrence of an event is 1.

We have data regarding the contribution of machines for manufacturing of bolts and the number of defective bolts manufactured. We have to calculate the probability of not getting a defective bolt from machine B given that it’s a defective bolt.
Let us denote the event of producing a bolt by machine A, B and C by A, B and C respectively. Let us denote the event of getting a defective bolt from machine A, B and C by X, Y and Z respectively. Let us denote the event of getting a defective by O.
We will calculate the probability of event O.
We will firstly evaluate the probability of each of the events.
We know that probability of any event is the ratio of number of favourable outcomes to total number of possible outcomes.
We know that the formula for conditional probability of occurrence of an event A given B has already occurred is given by $P\left( A|B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$.
Let’s assume that 100 bolts were manufactured by all the machines on a particular day.
We know that $x%$ of $y$ has the value $\dfrac{xy}{100}$.
Thus, the number of bolts manufactured by machine A $=\dfrac{30}{100}\left( 100 \right)=30$.
So, the probability of getting a bolt manufactured by machine A is $P\left( A \right)=\dfrac{30}{100}=0.3$.
The number of bolts manufactured by machine B $=\dfrac{50}{100}\left( 100 \right)=50$.
So, the probability of getting a bolt manufactured by machine B is $P\left( B \right)=\dfrac{50}{100}=0.5$.
The number of bolts manufactured by machine C $=\dfrac{20}{100}\left( 100 \right)=20$.
So, the probability of getting a bolt manufactured by machine C is $P\left( C \right)=\dfrac{20}{100}=0.2$.
We will now count the number of defective bolts produced by each machine.
We know that machine A produces $3%$ defective bolts out of the total number of bolts produced by machine A.
So, the number of defective bolts produced by machine A $=\dfrac{3}{100}\left( 30 \right)=0.9$.
Probability of producing a defective bolt by machine A $=P\left( X|A \right)=\dfrac{0.9}{30}=0.03$.
We know that machine B produces $4%$ defective bolts out of the total number of bolts produced by machine B.
So, the number of defective bolts produced by machine B $=\dfrac{4}{100}\left( 50 \right)=2$.
Probability of producing a defective bolt by machine B $=P\left( Y|B \right)=\dfrac{2}{50}=0.04$.
We know that machine C produces $1%$ defective bolts out of the total number of bolts produced by machine C.
So, the number of defective bolts produced by machine C $=\dfrac{1}{100}\left( 20 \right)=0.2$.
Probability of producing a defective bolt by machine C $=P\left( Z|C \right)=\dfrac{0.2}{20}=0.01$.
We will now calculate the probability of getting a defective bolt. A defective bolt can come from any three of the machines.
Thus, the probability of getting a defective bolt $=P\left( O \right)=P\left( X|A \right)P\left( A \right)+P\left( Y|B \right)P\left( B \right)+P\left( Z|C \right)P\left( C \right)$.
Thus, we have $P\left( O \right)=0.03\times 0.3+0.04\times 0.5+0.01\times 0.2=0.031$.
We will now calculate the probability of manufacturing a bolt from machine B, given that it’s a defected bolt, i.e., $P\left( B|O \right)$.
We can rewrite this as $P\left( B|O \right)=\dfrac{P\left( O|B \right)P\left( B \right)}{P\left( O \right)}$.
We know that $P\left( B \right)=0.5,P\left( O \right)=0.031$.
We will calculate the value of $P\left( O|B \right)$, which is the probability of getting a defective bolt given that machine B is manufacturing the bolts. We observe that $P\left( O|B \right)=P\left( Y\cap B \right)=P\left( Y|B \right)P\left( B \right)$.
Thus, we have $P\left( O|B \right)=P\left( Y\cap B \right)=P\left( Y|B \right)P\left( B \right)=0.04\times 0.5=0.02$.
So, we have $P\left( B|O \right)=\dfrac{P\left( O|B \right)P\left( B \right)}{P\left( O \right)}=\dfrac{0.02\times 0.5}{0.031}=0.322$.
We have to evaluate the probability of not manufacturing a bolt from machine B, given that it is a defective bolt, i.e., $P\left( {{B}^{c}}|O \right)=1-P\left( B|O \right)$.
Thus, we have $P\left( {{B}^{c}}|O \right)=1-P\left( B|O \right)=1-0.322=0.678$.
Hence, the probability of getting a defective bolt not manufactured by machine B is 0.678.

Note: We can also solve this question by assuming that the factory manufactures x bolts in a day and then calculate the probability of each of the events. It won’t affect the value of probability as probability simply represents a ratio. One must clearly know the definition of conditional probability, which is the occurrence of an event A given that event b has already occurred.

Last updated date: 22nd Sep 2023
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