Question

Of the three independent events \[{E_1}\], \[{E_2}\] and \[{E_3}\]​, the probability that only \[{E_1}\]​ occurs is \[\alpha \], only \[{E_2}\]​ occurs is \[\beta \] and only \[{E_3}\]​ occurs is \[\gamma \]. Let the probability \[p\] that none of events \[{E_1}\], \[{E_2}\] or \[{E_3}\] occurs satisfy the equations \[\left( {\alpha - {\rm{2}}\beta } \right){\rm{p}} = \alpha \beta \] and \[\left( {\beta - {\rm{3}}\gamma } \right){\rm{p}} = {\rm{2}}\beta \gamma \]. All the given probabilities are assumed to lie in the interval \[\left( {0,1} \right)\].
Then \[\dfrac{{{\rm{Probability\, of\, occurrence\, of\, }}{E_1}}}{{{\rm{Probability\, of\, occurrence\, of\, }}{E_3}}} = \]_____________

Answer 1

Hint: Here, we are required to find the fraction of probability of event \[{E_1}\] to that of event \[{E_3}\]. We will assume the probabilities of each event to be different variables. Then, we will find the values of \[\alpha \],\[\beta \], \[p\] and \[\gamma \] using the given information in the question. Substituting the values obtained in the given equations, and then substituting the value of \[y\] from the first equation in the second equation, we will be able to find the required fractions of the probability of occurrence of the required events.

Complete step-by-step answer:
Let the probability of event \[{E_1}\] be \[P\left( {{E_1}} \right) = x\]
Let the probability of event \[{E_2}\] be \[P\left( {{E_2}} \right) = y\]
And, let the probability of event \[{E_3}\] be \[P\left( {{E_3}} \right) = z\]
Now, according to the question, the probability that only \[{E_1}\] occurs is \[\alpha \].
Hence, the other two events can be written as \[\left( {1 - y} \right)\] and \[\left( {1 - z} \right)\] respectively.
Thus, we get,
\[x\left( {1 - y} \right)\left( {1 - z} \right) = \alpha \]
Similarly, it is given that the probability that only \[{E_2}\] occurs is \[\beta \]. Therefore,
\[\left( {1 - x} \right)y\left( {1 - z} \right) = \beta \]
And, the probability that only \[{E_3}\] occurs is \[\gamma \]. So,
\[\left( {1 - x} \right)\left( {1 - y} \right)z = \gamma \]
Also, according to the question \[p\] is the probability that none of the events occur.
Hence, we can write this as:
\[p = \left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right)\]
It is also given that \[\left( {\alpha - 2\beta } \right)p = \alpha \beta \].
Substituting the values of \[\alpha \],\[\beta \] and \[p\] in the above equation, we get,
\[\left[ {x\left( {1 - y} \right)\left( {1 - z} \right) - 2\left( {1 - x} \right)y\left( {1 - z} \right)} \right]\left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right) = xy\left( {1 - x} \right)\left( {1 - y} \right){\left( {1 - z} \right)^2}\]
Cancelling the same terms on both the sides, we get,
\[ \Rightarrow \left[ {x\left( {1 - y} \right)\left( {1 - z} \right) - 2\left( {1 - x} \right)y\left( {1 - z} \right)} \right] = xy\left( {1 - z} \right)\]
Factoring out common terms in LHS, we get
\[ \Rightarrow \left( {1 - z} \right)\left[ {x\left( {1 - y} \right) - 2\left( {1 - x} \right)y} \right] = xy\left( {1 - z} \right)\]
Again cancelling out the common term, we get
\[ \Rightarrow \left[ {x\left( {1 - y} \right) - 2\left( {1 - x} \right)y} \right] = xy\]
Now, opening the brackets, we get,
\[ \Rightarrow x - xy - 2y + 2xy = xy\]
\[ \Rightarrow x - 2y + xy = xy\]
Solving further, we get
\[ \Rightarrow x = 2y\]
Dividing both side by 2, we get
\[ \Rightarrow y = \dfrac{x}{2}\]……………………………………….\[\left( 1 \right)\]
Now, we are also given a second equation, \[\left( {\beta - 3\gamma } \right)p = 2\beta \gamma \].
Again substituting the values of all the variables, in the above equation, we get
\[\left[ {\left( {1 - x} \right)y\left( {1 - z} \right) - 3\left( {1 - x} \right)\left( {1 - y} \right)z} \right]\left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right) = 2yz{\left( {1 - x} \right)^2}\left( {1 - y} \right)\left( {1 - z} \right)\]
Cancelling the same terms on both the sides, we get,
\[ \Rightarrow \left[ {\left( {1 - x} \right)y\left( {1 - z} \right) - 3\left( {1 - x} \right)\left( {1 - y} \right)z} \right] = 2yz\left( {1 - x} \right)\]
Simplifying the expression, we get
\[ \Rightarrow \left[ {y\left( {1 - z} \right) - 3\left( {1 - y} \right)z} \right] = 2yz\]
\[ \Rightarrow y - yz - 3z + 3yz = 2yz\]
Subtracting the like terms, we get
\[ \Rightarrow y - 3z + 2yz = 2yz\]
Solving further, we get
\[ \Rightarrow y = 3z\]
\[ \Rightarrow \dfrac{y}{z} = 3\]
Now substituting \[y = \dfrac{x}{2}\] from \[\left( 1 \right)\], we get,
\[ \Rightarrow \dfrac{x}{{2z}} = 3\]
Multiplying both sides by 2, we get
\[ \Rightarrow \dfrac{x}{z} = \dfrac{6}{1}\]
We can rewrite this as:
\[ \Rightarrow \dfrac{{P\left( {{E_1}} \right)}}{{P\left( {{E_2}} \right)}} = \dfrac{6}{1}\]
Therefore, the required \[\dfrac{{{\rm{Probability\, of\, occurrence\, of \,}}{E_1}}}{{{\rm{Probability\, of\, occurrence\, of\, }}{E_2}}} = \dfrac{6}{1}\]
Hence, this is the required answer.

Note: Probability tells us the extent to which an event is likely to occur, i.e. the possibility of the occurrence of an event. Hence, it is measured by dividing the favorable outcomes by the total number of outcomes. Probability of an event varies from 0 to 1 that is it cannot be less than 0 or greater than 1. If the probability is 1 then the event is called a sure event, whereas the event with 0 probability never occurs.