
If the probability of an event A is given as \[P\left( A \right)=\dfrac{5}{9},\] then the odds against the event A is
(a) 5:9
(b) 5:4
(c) 4:5
(d) 5:14
Answer
602.7k+ views
Hint: As the value of P(A) is given to find what is asked, do use a formula which is \[P\left( O \right)=\dfrac{1-P\left( E \right)}{P\left( E \right)}\] where E is referred to the event which is A.
Complete step-by-step solution -
In the question, the value of P(A) is given which is \[\dfrac{5}{9}\] and we have to find the odds against event A. At first, we will know what probability i.e. Probability is the branch of mathematics concerning the numerical description of how likely an event is to occur or how likely it is that a proposition is true. Probability is a number between 0 and 1 where 0 indicates impossibility and 1 indicates certainty. The higher the probability of an event, the more likely it is that events will occur. A simple example is the tossing of a fair (unbiased) coin. Since the coin is fair, the outcomes are both equally probable “Heads” and “Tails”. The probability of heads equals the probability of tails and since no other outcomes are possible, the probability of either “heads” or “tails” is \[\dfrac{1}{2}\] (which can also be written as 0.5 or 50 %).
Now, for this question, we will use the formula,
\[P\left( O \right)=\dfrac{1-P\left( E \right)}{P\left( E \right)}\]
Here, E refers to the event and O refers to the odds against the event. In the question, E is A, so we can write it as,
\[P\left( O \right)=\dfrac{1-P\left( A \right)}{P\left( A \right)}\]
We are given the value of P(A) which is \[\dfrac{5}{9}.\]
So, the value of P(O) is
\[P\left( O \right)=\dfrac{1-P\left( A \right)}{P\left( A \right)}\]
\[\Rightarrow P\left( O \right)=\dfrac{1-\dfrac{5}{9}}{\dfrac{5}{9}}\]
\[\Rightarrow P\left( O \right)=\dfrac{\dfrac{4}{9}}{\dfrac{5}{9}}\]
\[\Rightarrow P\left( O \right)=\dfrac{4}{5}\]
So, we get the answer as 4:5.
Hence, the correct option is (c).
Note: In the question, if instead of odds against of event A, if odds favored was asked, then the formula would be \[\dfrac{P\left( E \right)}{1-P\left( E \right)}\] where E refers to the event.
Complete step-by-step solution -
In the question, the value of P(A) is given which is \[\dfrac{5}{9}\] and we have to find the odds against event A. At first, we will know what probability i.e. Probability is the branch of mathematics concerning the numerical description of how likely an event is to occur or how likely it is that a proposition is true. Probability is a number between 0 and 1 where 0 indicates impossibility and 1 indicates certainty. The higher the probability of an event, the more likely it is that events will occur. A simple example is the tossing of a fair (unbiased) coin. Since the coin is fair, the outcomes are both equally probable “Heads” and “Tails”. The probability of heads equals the probability of tails and since no other outcomes are possible, the probability of either “heads” or “tails” is \[\dfrac{1}{2}\] (which can also be written as 0.5 or 50 %).
Now, for this question, we will use the formula,
\[P\left( O \right)=\dfrac{1-P\left( E \right)}{P\left( E \right)}\]
Here, E refers to the event and O refers to the odds against the event. In the question, E is A, so we can write it as,
\[P\left( O \right)=\dfrac{1-P\left( A \right)}{P\left( A \right)}\]
We are given the value of P(A) which is \[\dfrac{5}{9}.\]
So, the value of P(O) is
\[P\left( O \right)=\dfrac{1-P\left( A \right)}{P\left( A \right)}\]
\[\Rightarrow P\left( O \right)=\dfrac{1-\dfrac{5}{9}}{\dfrac{5}{9}}\]
\[\Rightarrow P\left( O \right)=\dfrac{\dfrac{4}{9}}{\dfrac{5}{9}}\]
\[\Rightarrow P\left( O \right)=\dfrac{4}{5}\]
So, we get the answer as 4:5.
Hence, the correct option is (c).
Note: In the question, if instead of odds against of event A, if odds favored was asked, then the formula would be \[\dfrac{P\left( E \right)}{1-P\left( E \right)}\] where E refers to the event.
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