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Question Number 55583 by Tinkutara last updated on 27/Feb/19 | ||
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Answered by mr W last updated on 27/Feb/19 | ||
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$$\frac{\mathrm{2}×\mathrm{2}!\mathrm{2}!}{\mathrm{4}!}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\left({a}\right) \\ $$ | ||
Commented by mr W last updated on 27/Feb/19 | ||
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$$\left.{you}\:{are}\:{right}\:{sir}.\:{answer}\:{a}\right)\:{is}\:{correct}. \\ $$ | ||
Commented by Tinkutara last updated on 27/Feb/19 | ||
Thanks Sir! | ||
Commented by Tinkutara last updated on 27/Feb/19 | ||
Sir can you explain please? Why answer is not a because there may also be a case when two good machines are detected in first two trials? | ||