# Probability that at least one fails

A certain component of an electronic device has a probability of $0.1$ of failing. If there are $6$ such components in a circuit. What is the probability that at least one fails?

The Answer is $0.47$.

My Solution:

At least $1$ means more than $1$ failures

$P(1\, \text{fail}) = 0.1 \\ P(2\, \text{fails})=0.1\times0.1 \\P(3\, \text{fails})=0.1^{3}\\P(4\, \text{fails})=0.1^{4}\\P(5\, \text{fails})=0.1^{5}\\P(6\, \text{fails})=0.1^6\\ P(\text{Total})=P(1) +P(2)+...+P(6)=0.111111$

Where did I get wrong?

Among $6,1$ can fail in $\binom 61$ ways etc.

So, the required probability will be $$\sum_{1\le r\le 6}\binom 6 r\left(\frac1{10}\right)^r=1-\left(1-\frac1{10}\right)^6$$

Alternatively, the probability that at least one fails $=1-$ the probability that all succeed $$=1-\left(1-\frac1{10}\right)^6$$

• does that imply $P(All\ fails)=P(at\ least\ one\ fails)?$ because from the equation $P(all\ fail)=1-P(all\ succeed)$ is my analogy right?
– Jon
Mar 24, 2013 at 16:51
• @Jon, $P(A)+P(\bar A)=1$ where $\bar A$ implies the complement event of $A$. The complement event of at least one fail is "all succeed" Mar 24, 2013 at 17:33
• @labbhattacharjee The first method you used is some sort of binomial distribution. Shouldn't that be $\sum_{1\le r\lt 5}\binom 6 r\left(\frac1{10}\right)^r\left(\frac9{10}\right)^(6-r)=1-\left(1-\frac1{10}\right)^6$?
– Mick
Mar 11, 2014 at 16:37
• @Mick, wish I could read you expression. math.stackexchange.com/questions/697433/… Mar 11, 2014 at 16:39
• I am having some trouble in Latex. What I am saying is that you are using binomial distribution. What should have been done is Adding (1 bad 5 good), (2 bad 4 good), ... (5 bad 1 good) only in the form of SUM(r = 1 to 5 inclusive) of nCr 0.1^r[0.9]^[n-r] ?
– Mick
Mar 11, 2014 at 16:48

Probability of a component not to fail is 0.9. Therefore, the probability for 6 components not to fail (0.9)^6 = 0.531441.

1-0.531331 = 0.468559 or approximately 0.47.