A bag contains 10 red balls and 7 blue balls. A ball is drawn at random. The probability that ball drawn is not red is
(a) \[\dfrac{7}{10}\]
(b) \[\dfrac{3}{17}\]
(c) \[\dfrac{10}{17}\]
(d) \[\dfrac{7}{17}\]
Answer
663.6k+ views
Hint: In this question, as given that the ball which is picked at random should not be red we have only one chance that it should be blue. Now, we need to check the number of possible outcomes for the ball to be blue and the total number of outcomes possible. Then using the probability formula which is given by \[P=\dfrac{m}{n}\] which on further simplification gives the result.
Complete step-by-step answer:
PROBABILITY: If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by
\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}\]
Now, from the given conditions in the question we have 10 red balls and 7 blue balls
Now, the ball drawn at random to be not red it should be blue which is the only possible case
Here, the number of favourable outcomes for the ball to be blue is given by
\[m=7\]
Let us now find the total number of possible outcomes which is given by
\[n=10+7\]
Now, on further simplification we get,
\[n=17\]
Now, the probability for the ball to be drawn at random not red is given by
\[\Rightarrow P=\dfrac{m}{n}\]
Now, on further substituting the respective values we get,
\[\therefore P=\dfrac{7}{17}\]
Hence, the correct option is (d).
Note: Instead of finding the probability for the ball drawn at random to be blue we can also solve this by finding the probability for the ball drawn to be red and then subtracting if from 1 which also gives the same result as these events are complement to each other.It is important to note that while calculating the total number of possible outcomes we need to consider both colour balls if neglecting any one of them gives incorrect answer. It is also to be noted that respective values should be substituted accordingly because interchanging or incorrect substitution changes the result accordingly.
Complete step-by-step answer:
PROBABILITY: If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by
\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}\]
Now, from the given conditions in the question we have 10 red balls and 7 blue balls
Now, the ball drawn at random to be not red it should be blue which is the only possible case
Here, the number of favourable outcomes for the ball to be blue is given by
\[m=7\]
Let us now find the total number of possible outcomes which is given by
\[n=10+7\]
Now, on further simplification we get,
\[n=17\]
Now, the probability for the ball to be drawn at random not red is given by
\[\Rightarrow P=\dfrac{m}{n}\]
Now, on further substituting the respective values we get,
\[\therefore P=\dfrac{7}{17}\]
Hence, the correct option is (d).
Note: Instead of finding the probability for the ball drawn at random to be blue we can also solve this by finding the probability for the ball drawn to be red and then subtracting if from 1 which also gives the same result as these events are complement to each other.It is important to note that while calculating the total number of possible outcomes we need to consider both colour balls if neglecting any one of them gives incorrect answer. It is also to be noted that respective values should be substituted accordingly because interchanging or incorrect substitution changes the result accordingly.
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