Question

If E and F are independent events such that 0 < P(E) < 1 and 0 < P(F) < 1, then
A) E and F are mutually exclusive
B) E and F are independent
C) E‘ and F’ are independent
D) P(E|F) + P(E|F’) = 1

Answer 1

Hint:
Here, this question can be easily solved if one knows the relation between the independency of E’ and F’, and the independency of E and F’ when it has been given that E and F are independent events themselves.

Formula Used:
P(E|F) = \[\dfrac{{{\rm{Number of elementary}}\,{\rm{events favourable to E}} \cap {\rm{F }}}}{{{\rm{Number of elementary events favourable to F}}}}\]= \[\dfrac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}}\]

Complete step by step solution:
It has been given that E and F are independent events.
Which means,
P(E\[ \cap \]F) = P(E).P(F)
Thus, we can also say that E\[ \cap \]F and E\[ \cap \]F’ are mutually exclusive events.
Now, E = (E\[ \cap \]F) \[ \cup \] (E\[ \cap \]F’)
Which means, P(E) = P(E\[ \cap \]F) + P(E\[ \cap \]F’)
Shifting the terms,
P(E\[ \cap \]F’) = P(E) - P(E\[ \cap \]F)
P(E\[ \cap \]F’) = P(E) – P(E).P(F) (from above)
Taking common from right hand side of the equation, we get:
P(E\[ \cap \]F’) = P(E)(1-P(F))
P(E\[ \cap \]F’) = P(E).P(F’) …(i)
Hence, E and F’ are independent events.
Similarly, we can also show that E’ and F’ are independent events …(ii)
and we can also show that E’ and F are independent events.
Now, we have established the fact that E and F’ independent events, E’ and F’ are also independent events and finally that, E’ and F also independent events.
But this does not guarantee that E and F are mutually exclusive because it is not necessary that events have to be mutually exclusive for being independent.

Now, we know from the formula of conditional property of the given independent events E and F that:
P(E|F) = \[\dfrac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}}\] …(iii)
And similarly, we can apply the same formula for the conditional property of the above proven independent events E’ and F and we have got:
P(E|F’) = \[\dfrac{{P\left( {E \cap F'} \right)}}{{P\left( F \right)}}\] …(iv)
Now, adding equations (iii) and (iv) we have,
P(E|F) + P(E|F’) = \[\dfrac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}}\]+\[\dfrac{{P\left( {E \cap F'} \right)}}{{P\left( F \right)}}\]=\[\dfrac{{P\left( {E \cap F} \right) + P\left( {E \cap F'} \right)}}{{P\left( F \right)}}\]= \[\dfrac{{P(F)}}{{P(F)}}\]= \[1\] …(v)

Thus, from equations (i), (ii) and (v) we can say that options B, C and D are correct.

Note:
From this question we can observe that in order to avoid any mess, we must write the steps in order. It is also necessary that we correctly remember the relation between the events and their status (mutual exclusiveness, independence). And finally, it is crucial to accurately remember the formulae for all these things, as without them, nothing can be achieved.