Question #e8654

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Question #e8654

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Answer 1 · 2017-02-18T17:15:47

$P (both work) = \frac{1}{15}$ $P("both work") = 1/15$

$P (at least 1 fail) = \frac{14}{15}$ $P("at least 1 fail")=14/15$

Explanation:

Fractions are represented as parts of the whole of something. Consequently if you add up all the parts you should get a whole. The whole is expressed as 1. In that you may have 1 of a count of population of this country. So splitting the population into two equal counts is $\frac{1}{2}$ $1/2$ of the population. That is $\frac{1}{2}$ $1/2$ of 1 population.

So the sum of all probabilities is 1.
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Tony B Tony B

$P (Both work) = \frac{2}{6} \times \frac{1}{5} = \frac{1}{15}$ $P("Both work") = 2/6xx1/5=1/15$

$P (At least 1 fail) =$ $P("At least 1 fail") =$ 1 - probability all good $= 1 - \frac{1}{15} = \frac{14}{15}$ $= 1-1/15=14/15$

The logic hear is that if there is at least 1 bad then it is not possible to have all good. So you remove the all good $\to 1 - \frac{1}{15} = \frac{14}{15}$ $->1-1/15=14/15$

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Question #e8654

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