Question

Given that E and F are events such that \[P\left( E \right)=0.6,P\left( F \right)=0.3,P\left( E\cap F \right)=0.2\], find \[6P\left( F|E \right)\].

Answer 1

Hint: Use the formula for calculating the conditional probability of two given events which is \[P\left( F|E \right)=\dfrac{P\left( E\cap F \right)}{P\left( E \right)}\] and substitute the values of given probability of events.

We have two events \[E\] and \[F\] such that \[P\left( E \right)=0.6,P\left( F \right)=0.3,P\left( E\cap F \right)=0.2\]. We have to find the value of \[6P\left( F|E \right)\].
We will first evaluate the value of the conditional probability \[P\left( F|E \right)\] which is the probability of occurrence of event \[F\] given that the event \[E\] has already occurred.
We will use the formula for conditional probability which says that \[P\left( F|E \right)=\dfrac{P\left( E\cap F \right)}{P\left( E \right)}\].
Substituting the values \[P\left( E \right)=0.6,P\left( E\cap F \right)=0.2\] in the above formula, we get \[P\left( F|E \right)=\dfrac{P\left( E\cap F \right)}{P\left( E \right)}=\dfrac{0.2}{0.6}=\dfrac{2}{6}=\dfrac{1}{3}\].
Thus, we have \[P\left( F|E \right)=\dfrac{1}{3}\].
We now have to calculate \[6P\left( F|E \right)\]. Thus, we have \[6P\left( F|E \right)=6\left( \dfrac{1}{3} \right)=2\].
Hence, we have \[6P\left( F|E \right)=2\].
Probability of any event describes how likely an event is to occur or how likely it is that a proposition is true. The value of probability of any event always lies in the range \[\left[ 0,1 \right]\] where having \[0\] probability indicates that the event is impossible to happen, while having probability equal to \[1\] indicates that the event will surely happen. We must remember that the sum of probability of occurrence of some event and probability of non-occurrence of the same event is always \[1\].

Note: Conditional probability is a measure of the probability of occurrence of an event given that another event has occurred. \[P\left( A|B \right)\] measures the occurrence of event \[A\] given that event \[B\] has already occurred. If \[A\] and \[B\] are two independent events (which means that the probability of occurrence or non-occurrence of one event doesn’t affect the probability of occurring or non-occurring of the other event), then \[P\left( A|B \right)\] is simply the probability of occurrence of event \[A\], i.e. \[P\left( A \right)\].