
A box contains 5 red balls, 4 green balls and 7 white balls. A ball is drawn at random from the box. Find the probability that the drawn ball is neither red nor white?
( a ) $ \dfrac{5}{6} $
( b ) $ \dfrac{1}{4} $
( c ) $ \dfrac{1}{3} $
( d ) $ \dfrac{2}{7} $
Answer
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Hint: In this question, what we can do is we will first find out the total number of outcomes, total number of favourable outcomes and then by using the formula of finding the probability we can find the probability of getting a ball which is neither red nor white.
Complete step-by-step answer:
In question, it is given that there is a box and that box contains balls of three colours that are red, green and white. Number of red balls are 5, the number of green balls are 4 and the number of white balls are 7.
What we have to find is the probability of getting a ball which is neither red nor white when randomly a ball is taken out of a box.
Firstly, let us find the total number of balls in a box.
Total number of balls = number of red balls + number of green balls + number of white balls
Total number of balls = 5 + 4 + 7 = 16 balls
Now, the total number of events will be equal to the total number of balls in a box.
So, total number of events ( T ) = 16
In question it is asked that ball cannot be red or white, so number of unfavourable events will be equals to number of red balls + number of white balls that is
Total number of unfavourable events = 5 + 7 = 12
So, total number of favourable events = total number of events – total number of unfavourable events
So, total number of favourable events ( F ) = 16 – 12 = 4
We know that, \[\text{Probability=}\dfrac{\text{Total number of favourable outcome}}{\text{Total number of outcome}}\]
Hence, probability of not getting red ball as well as white ball will be $ =\dfrac{\text{total number of favourable events(F)}}{\text{total number of events(T)}} $
P ( getting neither red nor white ) = $ \dfrac{4}{16} $
On solving we get
P ( getting neither red nor white ) = $ \dfrac{1}{4} $
So, the correct answer is “Option B”.
Note: Favourable events are those events out of total events which have some outcome on trial and unfavourable outcomes are those which have no outcome on trial. Formula of finding probability must be known while solving the question.
Complete step-by-step answer:
In question, it is given that there is a box and that box contains balls of three colours that are red, green and white. Number of red balls are 5, the number of green balls are 4 and the number of white balls are 7.
What we have to find is the probability of getting a ball which is neither red nor white when randomly a ball is taken out of a box.
Firstly, let us find the total number of balls in a box.
Total number of balls = number of red balls + number of green balls + number of white balls
Total number of balls = 5 + 4 + 7 = 16 balls
Now, the total number of events will be equal to the total number of balls in a box.
So, total number of events ( T ) = 16
In question it is asked that ball cannot be red or white, so number of unfavourable events will be equals to number of red balls + number of white balls that is
Total number of unfavourable events = 5 + 7 = 12
So, total number of favourable events = total number of events – total number of unfavourable events
So, total number of favourable events ( F ) = 16 – 12 = 4
We know that, \[\text{Probability=}\dfrac{\text{Total number of favourable outcome}}{\text{Total number of outcome}}\]
Hence, probability of not getting red ball as well as white ball will be $ =\dfrac{\text{total number of favourable events(F)}}{\text{total number of events(T)}} $
P ( getting neither red nor white ) = $ \dfrac{4}{16} $
On solving we get
P ( getting neither red nor white ) = $ \dfrac{1}{4} $
So, the correct answer is “Option B”.
Note: Favourable events are those events out of total events which have some outcome on trial and unfavourable outcomes are those which have no outcome on trial. Formula of finding probability must be known while solving the question.
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