When the voltage is doubled, the electrical circuit may…

When the voltage is doubled, the electrical circuit may break because of the failure of one of three elements connected in series; the elements fail with probabilities $0.3$, $0.4$ and $0.6$ respectively. Determine the probability that the circuit will not break. By how much will this probability change if the first element is removed?

Step 1: Events

Let $A_k$ = "element $k$ works". Then $P(A_1)=1-0.3=0.7$, $P(A_2)=1-0.4=0.6$, $P(A_3)=1-0.6=0.4$. The circuit does not break iff all elements work: $A_1A_2A_3$.

Step 2: Multiplication rule

For independent events $P(A_1A_2A_3)=P(A_1)P(A_2)P(A_3)$:

\[ \boxed{P = 0.7\cdot0.6\cdot0.4 = 0.168.} \]

So the circuit breaks with probability $1-0.168=0.832$.

Step 3: Without the first element

\[ P' = 0.6\cdot0.4 = 0.24. \]

The probability that the circuit does not break increases by

\[ \boxed{P'-P = 0.24-0.168 = 0.072,} \]

i.e. it becomes $\frac{0.24}{0.168}=\frac1{0.7}\approx1.43$ times larger.

Step 4: Interpretation

In a series connection every extra element can only lower the reliability: the probability of no failure is multiplied by a number less than $1$. Removing an element with reliability $0.7$ divides the probability of failure-free work by $0.7$.