Consider two events E1 and E2 such that PE1-12- PE2-13 and PE1- E2-15- Which of the following statements is true-

Option (a) is wrong ∵ P(E_1∪ E_2)=12+13 15=1930≠23 Option (b) is wrong ∵ P(E_1∩ E_2)≠ P(E_1)P(E_2) Option (d) is wrong ∵ P(E_1/ E_2)=P(E_1∩ E_2)P(E_2 ...

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Engineering Mathematics

Conditional Probability

Consider two ...

Question

Consider two events $E_{1} a n d E_{2}$ such that $P (E_{1}) = \frac{1}{2}, P (E_{2}) = \frac{1}{3} a n d P (E_{1} \cap E_{2}) = \frac{1}{5} .$ Which of the following statements is true?

P (E_{1} \cup E_{2}) = \frac{2}{3}

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E_{1} a n d E_{2} a r e i n d e p e n d e n t

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E_{1} a n d E_{2} a r e n o t i n d e p e n d e n t

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

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Solution

The correct option is C $E_{1} a n d E_{2} a r e n o t i n d e p e n d e n t$
Option (a) is wrong

$∵ P (E_{1} \cup E_{2}) = \frac{1}{2} + \frac{1}{3} - \frac{1}{5} = \frac{19}{30} \neq \frac{2}{3}$

Option (b) is wrong

$∵ P (E_{1} \cap E_{2}) \neq P (E_{1}) P (E_{2})$

Option (d) is wrong

$∵ P (E_{1} / E_{2}) = \frac{P (E_{1} \cap E_{2})}{P (E_{2})} = \frac{\frac{1}{5}}{\frac{1}{3}} = \frac{3}{5} \neq \frac{4}{5}$

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Consider two events E1 and E2 such that P(E1)=12, P(E2)=13 and P(E1∩E2)=15. Which of the following statements is true?

The correct option is C E1 and E2 are not independentOption (a) is wrong ∵ P(E1∪E2)=12+13−15=1930≠23 Option (b) is wrong ∵ P(E1∩E2)≠P(E1)P(E2) Option (d) is wrong ∵ P(E1/E2)=P(E1∩E2)P(E2)=1513=35≠45

Consider two events $E_{1} a n d E_{2}$ such that $P (E_{1}) = \frac{1}{2}, P (E_{2}) = \frac{1}{3} a n d P (E_{1} \cap E_{2}) = \frac{1}{5} .$ Which of the following statements is true?