Question

Apiece of equipment will function only when all three parts A, B, C are working. The probability of part A failing during one year is 1/6 that of B failing is 1/20 and that of C failing is 1/10. What is the probability that the equipment will fail before the end of the year?​

Answer 1

The equipment fails means that atleast one among the parts A, B and C should fail. Because these three parts determine the function of the equipment.

So the probability that the equipment will fail, is equal to the probability that atleast one among the parts A, B and C will fail, whose complement is the probability that none among A, B and C will fail.

Probability that A will fail,

$\longrightarrow P(A')=\dfrac{1}{6}$

Probability that A won't fail,

$\longrightarrow P(A)=1-P(A')=1-\dfrac{1}{6}=\dfrac{5}{6}$

Probability that B will fail,

$\longrightarrow P(B')=\dfrac{1}{20}$

Probability that B won't fail,

$\longrightarrow P(B)=1-P(B')=1-\dfrac{1}{20}=\dfrac{19}{20}$

Probability that C will fail,

$\longrightarrow P(C')=\dfrac{1}{10}$

Probability that C won't fail,

$\longrightarrow P(C)=1-P(C')=1-\dfrac{1}{10}=\dfrac{9}{10}$

Now, probability that none among A, B and C will fail, or the equipment won't fail, is,

$\longrightarrow P(E)=P(A)\cdot P(B)\cdot P(C)$

$\longrightarrow P(E)=\dfrac{5}{6}\cdot\dfrac{19}{20}\cdot\dfrac{9}{10}$

$\longrightarrow P(E)=\dfrac{57}{80}$

Hence, the probability that the equipment will fail is,

$\longrightarrow P(E')=1-P(E)$

$\longrightarrow P(E')=1-\dfrac{57}{80}$

$\longrightarrow\underline{\underline{P(E')=\dfrac{23}{80}}}$

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• A piece of equipment will function only when all three parts A, B, C are working. The probability of part A failing during one year is 1/6 that of B failing is 1/20 and that of C failing is 1/10. What is the probability that the equipment will fail before the end of the year?

$\large\underline\purple{\bold{Solution :- }}$

According to statement,

Apiece of equipment will function only when all three parts A, B, C are working.

• The probability of part A fail is 1/6.

• The probability of part B fail is 1/20

• The probability of part C fail is 1/10.

Let X represents the event that part A works.

Let Y represents the event that part B works.

Let Z represents the event that part C works.

So,

$\rm : \implies \:P(X') = \: \dfrac{1}{6}$

This implies,

$\rm : \implies \:P(X) =1 - P(X') =1 - \dfrac{1}{6} = \dfrac{5}{6}$

Now,

$\rm : \implies \:P(Y') = \: \dfrac{1}{20}$

This implies,

$\rm : \implies \:P(Y) =1 - P(Y') =1 - \dfrac{1}{20} = \dfrac{19}{20}$

Now,

$\rm : \implies \:P(Z') =\dfrac{1}{10}$

This implies,

$\rm : \implies \:P(Z) =1 - P(Z') =1 - \dfrac{1}{10} = \dfrac{9}{10}$

Now,

• Probability that none of the component fails or equipment will work is given by

$\rm : \implies \:P(equipment \: works) \: = P(XYZ)$

$\rm : \implies \:P(equipment \: works) = P(X)P(Y)P(Z)$

$\rm : \implies \:P(equipment \: works) = \dfrac{5}{6} \times \dfrac{19}{20} \times \dfrac{9}{10}$

$\boxed{ \pink{\rm : \implies \:P(equipment \: works) = \dfrac{57}{80} }}$

Hence,

• The probability that the equipment will fail before the end of the year is given by

$\rm : \implies \:P(equipment \: not \: work) = \: 1 \: - P(equipment \: works)$

$\rm : \implies \:P(equipment \: not \: work) = \: 1 - \dfrac{57}{80}$

$\boxed{ \purple{\rm : \implies \:P(equipment \: not \: work) = \: \dfrac{23}{80} }}$