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
688.5k+ views
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. \[P\left( A \right)\].
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. \[P\left( A \right)\].
Recently Updated Pages
Which of the following graphs shows the variation of class 12 physics CBSE

Draw a labelled diagram of the human male reproductive class 12 biology CBSE

Describe the experiment to compare the emf of two cells class 12 physics CBSE

What is standard hydrogen electrode

What is conventional current and electric current class 12 physics CBSE

2Bromopentane is treated with an alcoholic KOH solution class 12 chemistry CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

A member of Simon commission later became Prime Minister class 12 social science CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Give one example of a reptile that is viviparous class 12 biology CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

