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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?

Jon
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2 Answers2

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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 $$=1-\left(1-\frac1{10}\right)^6$$

lab bhattacharjee
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  • 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 '13 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" – lab bhattacharjee Mar 24 '13 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 '14 at 16:37
  • @Mick, wish I could read you expression. http://math.stackexchange.com/questions/697433/a-family-has-three-children-what-is-the-probability-that-at-least-one-of-them-i – lab bhattacharjee Mar 11 '14 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 '14 at 16:48
  • @Mick, Have you gone through the link? You will get the same answer if you follow this tedious method. – lab bhattacharjee Mar 11 '14 at 16:51
  • Yes, I did. The alternative answer and the one on the RHS are correct but the LHS (ie. the tedious method) is not. – Mick Mar 11 '14 at 17:09
  • @labbhattacharjee I finally have the Latex fixed. If the question is solved by the tedious method (i.e. using the binomial method), it should be:- $$ \sum_{1\le r\le 6}\binom 6 r\left(\frac 1 {10}\right)^r\left(\frac 9 {10}\right)^{6-r}=1-\left(1-\frac 1 {10}\right)^6 $$ – Mick Mar 12 '14 at 13:55
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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.

libogista
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