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If E and F are events such that $P(E) = \dfrac{1}{4},P(F) = \dfrac{1}{2}{\text{ and P(E}} \cap {\text{F) = }}\dfrac{1}{8}$find:
$(1){\text{ P(E }} \cup {\text{ F)}}$
$(2){\text{ P}}\left( {\overline E \cap \overline F } \right)$Or $P\left( {{E^1} \cap {F^1}} \right)$

Last updated date: 29th May 2024
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Hint- Use the formulae for ${\text{P(E }} \cup {\text{ F)}}$ in the first part and use the concept of De Morgan’s law in the second part.

We have given, $P(E) = \dfrac{1}{4},P(F) = \dfrac{1}{2}{\text{ and P(E}} \cap {\text{F) = }}\dfrac{1}{8}$
$(1)$We know that${\text{P(E }} \cup {\text{ F)}} = P(E) + P(F) - P(E \cap F)$…………….. (1)
On substituting the values in equation (1) we get
${\text{P(E }} \cup {\text{ F)}} = \dfrac{1}{4} + \dfrac{1}{2} - \dfrac{1}{8} = \dfrac{{2 + 4 - 1}}{8} = \dfrac{5}{8}$
$(2)$Now from the solution of first part we know that
${\text{P(E }} \cup {\text{ F) = }}\dfrac{5}{8}$………………………………. (2)
Now using De Morgan’s law ${\left( {E \cup F} \right)^1} = \left( {{E^1} \cap {F^1}} \right)$
$ \Rightarrow P{\left( {E \cup F} \right)^1} = P\left( {{E^1} \cap {F^1}} \right)$……………………….. (3)
Now $P{\left( {E \cup F} \right)^1} = 1 - P\left( {E \cup F} \right)$ (As $P(\overline {E)} = 1 - P(E)$
So using equation (2)
$P{\left( {E \cup F} \right)^1} = 1 - \dfrac{5}{8} = \dfrac{3}{8}$
Using equation (3) we can say that
$ \Rightarrow $ $P\left( {{E^1} \cap {F^1}} \right) = \dfrac{3}{8}$

Note- Whenever we face such types of problems we need to have a good grasp of formula as these are mostly formula based only. Some of the important formulas and theorems are being stated above. The physical interpretation of ${\text{P(E }} \cup {\text{ F)}}$is that we need to find the probability of occurring of event E or F. The physical interpretation of ${\text{P}}\left( {\overline E \cap \overline F } \right)$is probability of neither event E occurring and nor event F occurring.

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