Question

A worker attends three machines each of which operates independently of the other two. The probabilities of the event that the machines will not require operator's intervention during a shift are equal to ${{p}_{1}}=0.4,{{p}_{2}}=0.3,{{p}_{3}}=0.2$. Find the probability of the event that at least one machine will require worker’s intervention during a shift.
A. $0.024$
B. $0.336$
C. $0.976$
D. $0.664$

Answer 1

Hint: We first find the condition which satisfies the event where at least one machine will require worker’s intervention during a shift. We find it as the complementary event of none of the machines requiring worker’s intervention during a shift.The independent events give us the probability. We subtract that from 1 to get the final solution.

Complete step by step answer:
A worker attends three machines each of which operates independently of the other two.The probabilities of the event that the machines will not require operator's intervention during a shift are equal to ${{p}_{1}}=0.4,{{p}_{2}}=0.3,{{p}_{3}}=0.2$. Now we find the probabilities of the event that the machines will require operator's intervention during a shift. We subtract them from 1 to get the values.So, we take the complementary events as
${{p}_{1}}^{c}=1-0.4=0.6 \\
\Rightarrow {{p}_{2}}^{c}=1-0.3=0.7 \\
\Rightarrow {{p}_{3}}^{c}=1-0.2=0.8 $
We have to find the probability of the event that at least one machine will require worker’s intervention during a shift. This will be the complementary event of none of the machines requiring worker’s intervention during a shift. This is equal to $P\left( {{p}_{1}}\cap {{p}_{2}}\cap {{p}_{3}} \right)$.

As the machines operate independently, we take the events as multiplication of their individual probabilities.Therefore,
$P\left( {{p}_{1}}\cap {{p}_{2}}\cap {{p}_{3}} \right)=P\left( {{p}_{1}} \right)\times P\left( {{p}_{2}} \right)\times P\left( {{p}_{3}} \right)$
Putting the values, we get
$P\left( {{p}_{1}}\cap {{p}_{2}}\cap {{p}_{3}} \right)=0.4\times 0.3\times 0.2 \\
\therefore P\left( {{p}_{1}}\cap {{p}_{2}}\cap {{p}_{3}} \right) =0.024$
Therefore, the probability of the event that at least one machine will require worker’s intervention during a shift is $1-0.024=0.976$.

Hence, the correct option is C.

Note: We can also denote the required event as the summation of multiple events where we need one, two or all three machines requiring worker’s intervention during a shift. We then use the complementary events of ${{p}_{1}}^{c},{{p}_{2}}^{c},{{p}_{3}}^{c}$.