
A box contains 9 red balls and 6 blue balls. Find the probability that the balls chosen at random are blue.
A.\[\dfrac{1}{5}\]
B.\[\dfrac{2}{5}\]
C.\[\dfrac{3}{5}\]
D.\[\dfrac{3}{15}\]
Answer
578.1k+ views
Hint: The probability of any given event is equal to the ratio of all the favourable outcomes with the total number of outcomes. It is the state of being probable. Here we will be using the given information and use the classical approach of probability.
Complete step-by-step answer:
Given: A box contains 9 red balls and 6 blue balls. We are to find the probability that the balls chosen at random are blue.
In this case, the box contains 9 red balls and 6 blue balls.
Hence, the sample space contains a total of 9+6=15 elements.
Therefore \[n\left( S \right) = 15\]
Again, let us define the event of drawing a blue ball as E.
Thus, the total number of ways this can be done is 6. So, the set of favourable outcomes contain 6 elements.
Therefore \[n\left( E \right) = 6\]
Now, the classical approach of probability defines this as the number of favourable outcomes/number of total outcomes.
Hence, the required probability is
\[\Rightarrow \dfrac{{n\left( E \right)}}{{n\left( S \right)}} = \dfrac{6}{{15}} = \dfrac{3}{5}\]
Note: In problems like these, it is to be always remembered that all the information given can be translated into concepts which can be solved using classical approach of probability. Always remember that the probability of any event lies between one and zero.
Complete step-by-step answer:
Given: A box contains 9 red balls and 6 blue balls. We are to find the probability that the balls chosen at random are blue.
In this case, the box contains 9 red balls and 6 blue balls.
Hence, the sample space contains a total of 9+6=15 elements.
Therefore \[n\left( S \right) = 15\]
Again, let us define the event of drawing a blue ball as E.
Thus, the total number of ways this can be done is 6. So, the set of favourable outcomes contain 6 elements.
Therefore \[n\left( E \right) = 6\]
Now, the classical approach of probability defines this as the number of favourable outcomes/number of total outcomes.
Hence, the required probability is
\[\Rightarrow \dfrac{{n\left( E \right)}}{{n\left( S \right)}} = \dfrac{6}{{15}} = \dfrac{3}{5}\]
Note: In problems like these, it is to be always remembered that all the information given can be translated into concepts which can be solved using classical approach of probability. Always remember that the probability of any event lies between one and zero.
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