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Importance of Brackets

If E= E 1 E...

Question

If $E = E_{1} E_{2} E_{3} E_{4} E_{5}$ and $P (E_{1}) = \frac{95}{100}$ , $P (E_{2} / E_{1}) = \frac{94}{99}$ , $P (E_{1} / E_{1} E_{2}) = \frac{93}{98}$ , $P (E_{4} / E_{1} E_{2} E_{3}) = \frac{92}{97}$ and $P (E_{5} / E_{1} E_{2} E_{3} E_{4}) = \frac{91}{96}$ , then find $P (E)$ .

A

P (E) = \frac{96.97.98.99.100}{91, 92.93.94.95}

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B

P (E) = \frac{91.92.93.94.95}{96, 97.98.99.100}

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C

P (E) = \frac{91.92.93.99.100}{96, 97.98.94.95}

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D

P (E) = \frac{91.92.93.94.100}{96, 97.98.99.95}

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Solution

The correct option is B $P (E) = \frac{91.92.93.94.95}{96, 97.98.99.100}$
$∵ P (E_{1} \cap E_{2} \cap E_{3} \cap E_{4} \cap E_{5})$ or $P (E_{1} E_{2} E_{3} E_{4} E_{5})$

$= P (E_{1}) . P (E_{2} / E_{1}) . P (E_{3} / E_{1} E_{2}) . P (E_{4} / E_{1} E_{2} E_{3}) . P (E_{5} / E_{1} E_{2} E_{3} E_{4})$

$∴ P (E) = \frac{95}{100} \times \frac{94}{99} \times \frac{93}{98} \times \frac{92}{97} \times \frac{91}{96}$
$= \frac{91.92.93.94.95}{96.97.98.99.100}$ $(∵ E = E_{1} E_{2} E_{3} E_{4} E_{5})$

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Similar questions

E = E_{1} \cap E_{2} \cap E_{3} \cap E_{4} \cap E_{5}

P (E_{1}) = \frac{95}{100}, P (E_{2} / E_{1}) = \frac{94}{99}, P (E_{3} / (E_{1} \cap E_{2})) = \frac{93}{98}, P (E_{4} / E_{1} \cap E_{2} \cap E_{3}) = \frac{92}{97}

P (E_{5} / E_{1} \cap E_{2} \cap E_{3} \cap E_{4}) = \frac{91}{96}

P (E)

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then P(E₄) =____________.

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E_{1}, E_{2}, E_{3} a n d E_{4}

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