In a bolt factory, three machines A, B and C manufacture 25%, 35% and 40% of the total production respectively. Of their respective outputs, 5%, 4% and 2% are defective. A bolt is drawn at random from the total production and it is found to be defective. What are the probabilities that it was manufactured by the machines A, B and C.?
A.$\dfrac{{25}}{{69}},\dfrac{{28}}{{69}},\dfrac{{16}}{{69}}$
B.$\dfrac{{25}}{{69}},\dfrac{{27}}{{69}},\dfrac{{17}}{{69}}$
C.$\dfrac{{28}}{{69}},\dfrac{{25}}{{69}},\dfrac{{16}}{{69}}$
D.$\dfrac{{27}}{{69}},\dfrac{{26}}{{69}},\dfrac{{16}}{{69}}$
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Hint: To find the probabilities of that a bolt is found to be defective was manufactured by the machine A, B and C we have to find the number of defective bolts produced by individual machines A, B and C and total number of defective bolts produced by all the machines. Take the total number of bolts produced in the bolt factory = 100.
Complete step-by-step answer:
Let us assume the total number of bolts produced is 100. Now we can find the number of bolts produced by machine A, B and C. As it is given that machine A, B and C produce 25%, 35% and 40% bolts respectively. So,
$\Rightarrow$ Total number of Bolts = 100
$\Rightarrow$ Machine A produces = $25\% of \, Total\, number\, of\, bolts = 25\% \times 100$
$\Rightarrow$ Machine B produces = $35\% of \, Total\, number\, of\, bolts = 35\% \times 100$
$ \Rightarrow$ Machine B produces = $\dfrac{{35}}{{100}} \times 100 = 35 bolts $
$ \Rightarrow$ Machine C produces = $40\% of \,Total\, number\, of\, bolts = 40\% \times 100$
$\Rightarrow$ Machine C produces = $\dfrac{{40}}{{100}} \times 100 = 40bolts$
Next step is to find the number of defective volts produced by machine A, B and C as it is given that 5%, 4% and 2% of bolts are defective respectively. So,
$
\Rightarrow Defective\, Bolts\, by\, Machine\, A = 5\% of Machine\, A\, produces = 5\% of\, 25 \\
\Rightarrow Defective\, Bolts\, by\, Machine\, A = \dfrac{5}{{100}} \times 25 = 1.25 \\
\\
\Rightarrow Defective\, Bolts\, by\, Machine\, B = 4\% of Machine\, B\, produces = 4\% of\, 35 \\
\Rightarrow Defective\, Bolts\, by\, Machine\, B = \dfrac{4}{{100}} \times 35 = 1.40 \\
\\
\Rightarrow Defective\, Bolts\, by\, Machine\, C = 2\% of\, Machine\, C\, produces = 2\% of\, 40 \\
\Rightarrow Defective\, Bolts\, by\, Machine\, C = \dfrac{2}{{100}} \times 40 = 0.8 \\
$
Let P(A) is the probability of defective bolt produced by machine A, Let P(B) is the probability of defective bolt produced by machine B and Let P(C) is the probability of defective bolt produced by machine C. Probability is given by
$
\Rightarrow P(n) = \dfrac{{Number\, of\, Favourable\, Events}}{{Total\, Number\, of\, Events}} \\
\\
\Rightarrow Tota\operatorname{l} Number\, of\, Events = 1.25 + 1.40 + 0.8 = 3.45 \\
\Rightarrow P(A) = \dfrac{{1.25}}{{3.45}} = \dfrac{{25}}{{69}} \\
\Rightarrow P(B) = \dfrac{{1.40}}{{3.45}} = \dfrac{{28}}{{69}} \\
\Rightarrow P(C) = \dfrac{{0.8}}{{3.45}} = \dfrac{{16}}{{69}} \\
$
So, the correct option is A.
Note: We can also solve this question orally which saves time during exams as it is a multiple choice question. We can see through multiplying 25 by 5, 35 by 4 and 40 by 2 that machine B has maximum outcome of defective bolts. So, C and D are wrong and there is only A and B. Now check that 25 factors are 5 and 5, 35 factors include 7 and 16 includes factors 4 & 2. So, option A is correct.
Complete step-by-step answer:
Let us assume the total number of bolts produced is 100. Now we can find the number of bolts produced by machine A, B and C. As it is given that machine A, B and C produce 25%, 35% and 40% bolts respectively. So,
$\Rightarrow$ Total number of Bolts = 100
$\Rightarrow$ Machine A produces = $25\% of \, Total\, number\, of\, bolts = 25\% \times 100$
$\Rightarrow$ Machine B produces = $35\% of \, Total\, number\, of\, bolts = 35\% \times 100$
$ \Rightarrow$ Machine B produces = $\dfrac{{35}}{{100}} \times 100 = 35 bolts $
$ \Rightarrow$ Machine C produces = $40\% of \,Total\, number\, of\, bolts = 40\% \times 100$
$\Rightarrow$ Machine C produces = $\dfrac{{40}}{{100}} \times 100 = 40bolts$
Next step is to find the number of defective volts produced by machine A, B and C as it is given that 5%, 4% and 2% of bolts are defective respectively. So,
$
\Rightarrow Defective\, Bolts\, by\, Machine\, A = 5\% of Machine\, A\, produces = 5\% of\, 25 \\
\Rightarrow Defective\, Bolts\, by\, Machine\, A = \dfrac{5}{{100}} \times 25 = 1.25 \\
\\
\Rightarrow Defective\, Bolts\, by\, Machine\, B = 4\% of Machine\, B\, produces = 4\% of\, 35 \\
\Rightarrow Defective\, Bolts\, by\, Machine\, B = \dfrac{4}{{100}} \times 35 = 1.40 \\
\\
\Rightarrow Defective\, Bolts\, by\, Machine\, C = 2\% of\, Machine\, C\, produces = 2\% of\, 40 \\
\Rightarrow Defective\, Bolts\, by\, Machine\, C = \dfrac{2}{{100}} \times 40 = 0.8 \\
$
Let P(A) is the probability of defective bolt produced by machine A, Let P(B) is the probability of defective bolt produced by machine B and Let P(C) is the probability of defective bolt produced by machine C. Probability is given by
$
\Rightarrow P(n) = \dfrac{{Number\, of\, Favourable\, Events}}{{Total\, Number\, of\, Events}} \\
\\
\Rightarrow Tota\operatorname{l} Number\, of\, Events = 1.25 + 1.40 + 0.8 = 3.45 \\
\Rightarrow P(A) = \dfrac{{1.25}}{{3.45}} = \dfrac{{25}}{{69}} \\
\Rightarrow P(B) = \dfrac{{1.40}}{{3.45}} = \dfrac{{28}}{{69}} \\
\Rightarrow P(C) = \dfrac{{0.8}}{{3.45}} = \dfrac{{16}}{{69}} \\
$
So, the correct option is A.
Note: We can also solve this question orally which saves time during exams as it is a multiple choice question. We can see through multiplying 25 by 5, 35 by 4 and 40 by 2 that machine B has maximum outcome of defective bolts. So, C and D are wrong and there is only A and B. Now check that 25 factors are 5 and 5, 35 factors include 7 and 16 includes factors 4 & 2. So, option A is correct.