Himanshuv0 wrote:
Hello
Bunuel,
I don't know where my reasoning is faltering. Can you please help me in locating the falter ?
My reasoning is that if any one of lightbulb fails, then the entire string will fails. And Probability of failing one light bulb is given the question stem. Therefore P(one lightbulb failing) = P (entire string failing).
So if it is 100 bulb string or 10 bulb string, the string will fail if any one of the bulb fails.
Himanshuv0: you wrote "P(one lightbulb failing) = P (entire string failing)". This is false, because the probability of failure goes up when we are evaluating the entire string, instead of just one individual light bulb. The logic is that there are more opportunities for failure in the entire string. Similarly, our odds of winning a lottery go up if we buy more lottery tickets.
Or, let's illustrate with a dice game example:
Suppose we lose if we roll a "1" on any of the dice.
If we roll just 1 dice, our chances of losing are \(\frac{1}{6}\).
However, what are the chances of losing if we roll 3 dice? Our chances of losing increase, because there are many possible losing combinations for the 3 dice:
ONN (One, Not One, Not One)
NON
NNO
OON
ONO
NOO
OOO
So, one way to solve would be to add all of those probabilities up.
However, a much easier way is using the formula: (1 - Probability of NOT a loss).
The probability of NOT a loss is: \(\frac{5}{6} * \frac{5}{6} * \frac{5}{6} = \frac{125}{216}\)
Therefore, the probability of a loss is \(1 - \frac{125}{216} = \frac{91}{216}\)
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