# A box contains 4 white balls, 6 red balls, 7 black balls and 3 blue balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is :

A. neither white nor black

B. red or white, and

C. either white or red or black or blue

Last updated date: 20th Mar 2023

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Answer

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Hint: Use the fundamental definition of probability that it can be measured by ratio of number of favorable cases to total cases/ sample spaces.

Complete step-by-step answer:

Here, it is given that Box contains 4 white balls, 6 red balls, 7 black balls and 3 blue balls i.e. in total 20 balls are present in the box.

Hence, Total balls in Box = 20.

A. neither white nor black

Probability of drawing neither white nor black balls means drawing red balls or blue balls.

So, total probability of drawing Red balls or Blue balls = Probability of drawing Red balls +

Probability of drawing Blue balls.

Now, we know the definition of probability of an event can be given as

P(E)=Number of favorable cases/Total cases\[\ldots \ldots (1)\]

Hence, Probability of drawing Red balls or Blue balls=

$P=\dfrac{6}{20}+\dfrac{3}{20}=\dfrac{9}{20}$

Therefore, Probability of drawing neither white nor black be$\dfrac{9}{20}$.

B. Red or white

Probability of drawing either Red of White balls can be given as a summation of probability

of drawing Red Balls and white balls both.

Number of Red balls= 6

Number of white balls= 4

Total number of balls= 20

Now, from equation (1) , we can give required probability as

P=Probability of drawing Red balls + Probability of drawing white balls

Hence, we get

$P=\dfrac{6}{20}+\dfrac{4}{20}=\dfrac{10}{20}=\dfrac{1}{2}$

C. either white or red or black or blue

Here, we need to involve the probability of drawing each colour as we are drawing either of

all balls of any colour.

So, here probability of either white or red or black or blue can be given as

P= Probability of drawing (Red)+(White)+(Black)+(Blue)

As we have

Number of Red balls= 6

Number of White balls= 4

Number of black balls= 7

Number of blue balls= 3

Total balls=20

Hence,

$P=\dfrac{4}{20}+\dfrac{6}{20}+\dfrac{7}{20}+\dfrac{3}{20}=\dfrac{20}{20}=1$

Therefore, answers of the problem are $\dfrac{9}{20},\dfrac{1}{2},1.$

Note: One can miss any of the colour while writing the probability of drawing that colour. He/she may involve favourable cases of any other colour ball.

First case can also be calculated by getting the probability of either white or black balls then subtract it from 1 as $1-\dfrac{1}{20}=\dfrac{19}{20}$.

One can go wrong while writing probability of red or white (2nd case) or with the last part as well. One can multiply the probability of red and white balls to get probability of red or white which is the wrong approach. So, take care of it in these kinds of questions.

Complete step-by-step answer:

Here, it is given that Box contains 4 white balls, 6 red balls, 7 black balls and 3 blue balls i.e. in total 20 balls are present in the box.

Hence, Total balls in Box = 20.

A. neither white nor black

Probability of drawing neither white nor black balls means drawing red balls or blue balls.

So, total probability of drawing Red balls or Blue balls = Probability of drawing Red balls +

Probability of drawing Blue balls.

Now, we know the definition of probability of an event can be given as

P(E)=Number of favorable cases/Total cases\[\ldots \ldots (1)\]

Hence, Probability of drawing Red balls or Blue balls=

$P=\dfrac{6}{20}+\dfrac{3}{20}=\dfrac{9}{20}$

Therefore, Probability of drawing neither white nor black be$\dfrac{9}{20}$.

B. Red or white

Probability of drawing either Red of White balls can be given as a summation of probability

of drawing Red Balls and white balls both.

Number of Red balls= 6

Number of white balls= 4

Total number of balls= 20

Now, from equation (1) , we can give required probability as

P=Probability of drawing Red balls + Probability of drawing white balls

Hence, we get

$P=\dfrac{6}{20}+\dfrac{4}{20}=\dfrac{10}{20}=\dfrac{1}{2}$

C. either white or red or black or blue

Here, we need to involve the probability of drawing each colour as we are drawing either of

all balls of any colour.

So, here probability of either white or red or black or blue can be given as

P= Probability of drawing (Red)+(White)+(Black)+(Blue)

As we have

Number of Red balls= 6

Number of White balls= 4

Number of black balls= 7

Number of blue balls= 3

Total balls=20

Hence,

$P=\dfrac{4}{20}+\dfrac{6}{20}+\dfrac{7}{20}+\dfrac{3}{20}=\dfrac{20}{20}=1$

Therefore, answers of the problem are $\dfrac{9}{20},\dfrac{1}{2},1.$

Note: One can miss any of the colour while writing the probability of drawing that colour. He/she may involve favourable cases of any other colour ball.

First case can also be calculated by getting the probability of either white or black balls then subtract it from 1 as $1-\dfrac{1}{20}=\dfrac{19}{20}$.

One can go wrong while writing probability of red or white (2nd case) or with the last part as well. One can multiply the probability of red and white balls to get probability of red or white which is the wrong approach. So, take care of it in these kinds of questions.

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