SATHEE: Probability 3 Question 28

Probability 3 Question 28

28. If $E$ and $F$ are independent events such that $0 < P (E) < 1$ and $0 < P (F) < 1$ , then

$(1989, 2 M)$

(a) $E$ and $F$ are mutually exclusive

(b) $E$ and $F^{c}$ (the complement of the event $F$ ) are independent

(d) $P (E / F) + P (E^{c} / F) = 1$

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Solution:

Since, $E$ and $F$ are independent events. Therefore, $P (E \cap F) = P (E) \cdot P (F) \neq 0$ , so $E$ and $F$ are not mutually exclusive events.

Now, $P (E \cap \bar{F}) = P (E) - P (E \cap F) = P (E) - P (E) \cdot P (F)$

$= P (E) [1 - P (F)] = P (E) \cdot P (\bar{F})$

and $P (\bar{E} \cap \bar{F}) = P (\overset{―}{E \cup F}) = 1 - P (E \cup F)$

$\begin{aligned} = 1 - [1 - P (\bar{E}) \cdot P (\bar{F})] \\ = P (\bar{E}) \cdot P (\bar{F}) \end{aligned}$

So, $E$ and $\bar{F}$ as well as $\bar{E}$ and $\bar{F}$ are independent events.

Now, $P (E / F) + P (\bar{E} / F) = \frac{P (E \cap F) + P (\bar{E} \cap F)}{P (F)}$

$= \frac{P (F)}{P (F)} = 1$

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Probability 3 Question 28

28. If $E$ and $F$ are independent events such that $0 < P (E) < 1$ and $0 < P (F) < 1$ , then

Fill in the Blanks

Solution:

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चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

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एनसीईआरटी व्यायाम वीडियो समाधान

Probability 3 Question 28

28. If E and F are independent events such that 0<P(E)<1 and 0<P(F)<1, then

Fill in the Blanks

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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28. If $E$ and $F$ are independent events such that $0 < P (E) < 1$ and $0 < P (F) < 1$ , then