If E- E1- E2- E3- E4- E5 and PE1- 95100- PE2- E1 - 9499- PE3-E1- E2-9398- PE4-E1- E2- E3-9297 and PE5-E1- E2- E3- E4-9196- then PE is-

We know, nbsp; 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) Given: E=E_1 ∩ E_2∩ E_3∩ E_4∩ ...

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Standard VII

Mathematics

Complement of Set

If E= E1∩ E2∩...

Question

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

P (E) = \frac{96 \cdot 97 \cdot 98 \cdot 99 \cdot 100}{91 \cdot 92 \cdot 93 \cdot 94 \cdot 95}

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P (E) = \frac{91 \cdot 92 \cdot 93 \cdot 94 \cdot 95}{96 \cdot 97 \cdot 98 \cdot 99 \cdot 100}

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P (E) = \frac{91 \cdot 92 \cdot 93}{96 \cdot 97 \cdot 98}

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P (E) = \frac{91 \cdot 92 \cdot 93 \cdot 94}{96 \cdot 97 \cdot 98 \cdot 99}

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Solution

The correct option is B $P (E) = \frac{91 \cdot 92 \cdot 93 \cdot 94 \cdot 95}{96 \cdot 97 \cdot 98 \cdot 99 \cdot 100}$
We know,
$P (E_{1} \cap E_{2} \cap E_{3} \cap E_{4} \cap E_{5}) = P (E_{1}) . P (E_{2} / E_{1}) . P (E_{3} / E_{1} \cap E_{2}) . P (E_{4} / E_{1} \cap E_{2} \cap E_{3}) . P (E_{5} / E_{1} \cap E_{2} \cap E_{3} \cap E_{4})$
Given: $E = E_{1} \cap E_{2} \cap E_{3} \cap E_{4} \cap E_{5}$
$∴ P (E) = \frac{95}{100} \times \frac{94}{99} \times \frac{93}{98} \times \frac{92}{97} \times \frac{91}{96}$
$\Rightarrow P (E) = \frac{91 \cdot 92 \cdot 93 \cdot 94 \cdot 95}{96 \cdot 97 \cdot 98 \cdot 99 \cdot 100}$

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Similar questions

Q. If

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

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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}

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If E=E1∩E2∩E3∩E4∩E5 and P(E1)=95100,P(E2/E1)=9499,P(E3/(E1∩E2))=9398,P(E4/E1∩E2∩E3)=9297 and P(E5/E1∩E2∩E3∩E4)=9196, then P(E) is: