# A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:

$(a)$ White

$(b)$ Red

$(c)$ Black

$(d)$ Not red

Last updated date: 25th Mar 2023

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Answer

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Hint: This bag contains 3 red, 5 black and 4 white balls. Now one ball is drawn at random so use the concept of probability that it is a favorable outcome divided by the total outcome depending upon the requirement of the question.

Total number of bags in the bag is 12 (3 red balls, 5 black balls, 4 white balls)

$(a)$ White -

Let A be the event of drawing a white ball.

Now in total we have 4 white balls, thus the number of ways of selecting one white out of 4 white ball is $^4{C_1}$

Thus favorable outcomes for event A is$^4{C_1}$……………………… (1)

Now there are in total 12 balls in the bag, so the ways of selecting one ball out of these 12 balls will be$^{12}{C_1}$.

Thus the total number of possible outcomes for the event A is $^{12}{C_1}$…………………. (2)

Now probability of an event A that is ${\text{P(A) = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcome}}}}$……………………… (3)

Thus substituting the values from equation (1) and equation (2) in equation (3) we get

${\text{P(A) = }}\dfrac{{^4{C_1}}}{{^{12}{C_1}}}$

Now using the formula of$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, let’s calculate the value of $\dfrac{{^4{C_1}}}{{^{12}{C_1}}}$hence it will be

$ \Rightarrow \dfrac{{\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}}}}{{\dfrac{{12!}}{{1!\left( {12 - 1} \right)!}}}} = \dfrac{{\dfrac{{4!}}{{3!}}}}{{\dfrac{{12!}}{{11!}}}}$………………………………… (4)

Now using $n! = n \times (n - 1) \times (n - 2) \times (n - 3)...........(n - r){\text{ r < n}}$equation (4) can be simplified as

$ \Rightarrow \dfrac{{\dfrac{{4 \times 3!}}{{3!}}}}{{\dfrac{{12 \times 11!}}{{11!}}}} = \dfrac{4}{{12}} = \dfrac{1}{3}$

Hence the probability of drawing a white ball from the bag containing 3 red, 5 black and 4 white balls is$\dfrac{1}{3}$.

$(b)$ Red ball-

Let B be the event of drawing a red ball.

Now in total we have 3 red balls, thus the number of ways of selecting one red out of 3 red balls is $^3{C_1}$

Thus favorable outcomes for event B is$^3{C_1}$……………………… (5)

Now there are in total 12 balls in the bag, so the ways of selecting one ball out of these 12 balls will be$^{12}{C_1}$.

Thus the total number of possible outcomes for the event B is $^{12}{C_1}$…………………. (6)

Now probability of an event A that is ${\text{P(B) = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcome}}}}$……………………… (7)

Thus substituting the values from equation (5) and equation (6) in equation (7) we get

${\text{P(B) = }}\dfrac{{^3{C_1}}}{{^{12}{C_1}}}$

Now using the formula of$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, let’s calculate the value of $\dfrac{{^3{C_1}}}{{^{12}{C_1}}}$hence it will be

$ \Rightarrow \dfrac{{\dfrac{{3!}}{{1!\left( {3 - 1} \right)!}}}}{{\dfrac{{12!}}{{1!\left( {12 - 1} \right)!}}}} = \dfrac{{\dfrac{{3!}}{{2!}}}}{{\dfrac{{12!}}{{11!}}}}$………………………………… (8)

Now using $n! = n \times (n - 1) \times (n - 2) \times (n - 3)...........(n - r){\text{ r < n}}$equation (8) can be simplified as

$ \Rightarrow \dfrac{{\dfrac{{3 \times 2!}}{{2!}}}}{{\dfrac{{12 \times 11!}}{{11!}}}} = \dfrac{3}{{12}} = \dfrac{1}{4}$

Hence the probability of drawing a red ball from the bag containing 3 red, 5 black and 4 white balls is$\dfrac{1}{4}$.

$(c)$ Black ball-

Let C be the event of drawing a black ball.

Now in total we have 5 black balls, thus the number of ways of selecting one black ball out of 5 black balls is $^5{C_1}$

Thus favorable outcomes for event C is$^5{C_1}$……………………… (9)

Now there are in total 12 balls in the bag, so the ways of selecting one ball out of these 12 balls will be$^{12}{C_1}$.

Thus the total number of possible outcomes for the event C is $^{12}{C_1}$…………………. (10)

Now probability of an event A that is ${\text{P(C) = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcome}}}}$……………………… (11)

Thus substituting the values from equation (5) and equation (6) in equation (7) we get

${\text{P(C) = }}\dfrac{{^5{C_1}}}{{^{12}{C_1}}}$

Now using the formula of$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, let’s calculate the value of $\dfrac{{^5{C_1}}}{{^{12}{C_1}}}$hence it will be

$ \Rightarrow \dfrac{{\dfrac{{5!}}{{1!\left( {5 - 1} \right)!}}}}{{\dfrac{{12!}}{{1!\left( {12 - 1} \right)!}}}} = \dfrac{{\dfrac{{5!}}{{4!}}}}{{\dfrac{{12!}}{{11!}}}}$………………………………… (12)

Now using $n! = n \times (n - 1) \times (n - 2) \times (n - 3)...........(n - r){\text{ r < n}}$equation (8) can be simplified as

$ \Rightarrow \dfrac{{\dfrac{{5 \times 4!}}{{4!}}}}{{\dfrac{{12 \times 11!}}{{11!}}}} = \dfrac{5}{{12}}$

Hence the probability of drawing a black ball from the bag containing 3 red, 5 black and 4 white balls is$\dfrac{5}{{12}}$.

$(d)$ Non red ball -

Let D be the event of drawing a non-red ball.

Now in total we have 9 non-red balls (5 black and 4 white), thus the number of ways of selecting one non-red ball out of 9 non-red ball is $^9{C_1}$

Thus favorable outcomes for event D is$^9{C_1}$……………………… (13)

Now there are in total 12 balls in the bag, so the ways of selecting one ball out of these 12 balls will be$^{12}{C_1}$.

Thus the total number of possible outcomes for the event D is $^{12}{C_1}$…………………. (14)

Now probability of an event A that is ${\text{P(D) = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcome}}}}$……………………… (15)

Thus substituting the values from equation (13) and equation (14) in equation (15) we get

${\text{P(D) = }}\dfrac{{^9{C_1}}}{{^{12}{C_1}}}$

Now using the formula of$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, let’s calculate the value of $\dfrac{{^9{C_1}}}{{^{12}{C_1}}}$hence it will be

$ \Rightarrow \dfrac{{\dfrac{{9!}}{{1!\left( {9 - 1} \right)!}}}}{{\dfrac{{12!}}{{1!\left( {12 - 1} \right)!}}}} = \dfrac{{\dfrac{{9!}}{{8!}}}}{{\dfrac{{12!}}{{11!}}}}$………………………………… (12)

Now using $n! = n \times (n - 1) \times (n - 2) \times (n - 3)...........(n - r){\text{ r < n}}$equation (8) can be simplified as

$ \Rightarrow \dfrac{{\dfrac{{9 \times 8!}}{{8!}}}}{{\dfrac{{12 \times 11!}}{{11!}}}} = \dfrac{9}{{12}} = \dfrac{3}{4}$

Hence the probability of drawing a non-red ball from the bag containing 3 red, 5 black and 4 white balls is$\dfrac{3}{4}$.

Note: Whenever we face such types of problems the key concept that we need to apply is simply to find the number of favorable outcomes to that particular event and then the total possible outcome for that particular event. This will help get the right answer to the problem.

Total number of bags in the bag is 12 (3 red balls, 5 black balls, 4 white balls)

$(a)$ White -

Let A be the event of drawing a white ball.

Now in total we have 4 white balls, thus the number of ways of selecting one white out of 4 white ball is $^4{C_1}$

Thus favorable outcomes for event A is$^4{C_1}$……………………… (1)

Now there are in total 12 balls in the bag, so the ways of selecting one ball out of these 12 balls will be$^{12}{C_1}$.

Thus the total number of possible outcomes for the event A is $^{12}{C_1}$…………………. (2)

Now probability of an event A that is ${\text{P(A) = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcome}}}}$……………………… (3)

Thus substituting the values from equation (1) and equation (2) in equation (3) we get

${\text{P(A) = }}\dfrac{{^4{C_1}}}{{^{12}{C_1}}}$

Now using the formula of$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, let’s calculate the value of $\dfrac{{^4{C_1}}}{{^{12}{C_1}}}$hence it will be

$ \Rightarrow \dfrac{{\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}}}}{{\dfrac{{12!}}{{1!\left( {12 - 1} \right)!}}}} = \dfrac{{\dfrac{{4!}}{{3!}}}}{{\dfrac{{12!}}{{11!}}}}$………………………………… (4)

Now using $n! = n \times (n - 1) \times (n - 2) \times (n - 3)...........(n - r){\text{ r < n}}$equation (4) can be simplified as

$ \Rightarrow \dfrac{{\dfrac{{4 \times 3!}}{{3!}}}}{{\dfrac{{12 \times 11!}}{{11!}}}} = \dfrac{4}{{12}} = \dfrac{1}{3}$

Hence the probability of drawing a white ball from the bag containing 3 red, 5 black and 4 white balls is$\dfrac{1}{3}$.

$(b)$ Red ball-

Let B be the event of drawing a red ball.

Now in total we have 3 red balls, thus the number of ways of selecting one red out of 3 red balls is $^3{C_1}$

Thus favorable outcomes for event B is$^3{C_1}$……………………… (5)

Now there are in total 12 balls in the bag, so the ways of selecting one ball out of these 12 balls will be$^{12}{C_1}$.

Thus the total number of possible outcomes for the event B is $^{12}{C_1}$…………………. (6)

Now probability of an event A that is ${\text{P(B) = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcome}}}}$……………………… (7)

Thus substituting the values from equation (5) and equation (6) in equation (7) we get

${\text{P(B) = }}\dfrac{{^3{C_1}}}{{^{12}{C_1}}}$

Now using the formula of$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, let’s calculate the value of $\dfrac{{^3{C_1}}}{{^{12}{C_1}}}$hence it will be

$ \Rightarrow \dfrac{{\dfrac{{3!}}{{1!\left( {3 - 1} \right)!}}}}{{\dfrac{{12!}}{{1!\left( {12 - 1} \right)!}}}} = \dfrac{{\dfrac{{3!}}{{2!}}}}{{\dfrac{{12!}}{{11!}}}}$………………………………… (8)

Now using $n! = n \times (n - 1) \times (n - 2) \times (n - 3)...........(n - r){\text{ r < n}}$equation (8) can be simplified as

$ \Rightarrow \dfrac{{\dfrac{{3 \times 2!}}{{2!}}}}{{\dfrac{{12 \times 11!}}{{11!}}}} = \dfrac{3}{{12}} = \dfrac{1}{4}$

Hence the probability of drawing a red ball from the bag containing 3 red, 5 black and 4 white balls is$\dfrac{1}{4}$.

$(c)$ Black ball-

Let C be the event of drawing a black ball.

Now in total we have 5 black balls, thus the number of ways of selecting one black ball out of 5 black balls is $^5{C_1}$

Thus favorable outcomes for event C is$^5{C_1}$……………………… (9)

Now there are in total 12 balls in the bag, so the ways of selecting one ball out of these 12 balls will be$^{12}{C_1}$.

Thus the total number of possible outcomes for the event C is $^{12}{C_1}$…………………. (10)

Now probability of an event A that is ${\text{P(C) = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcome}}}}$……………………… (11)

Thus substituting the values from equation (5) and equation (6) in equation (7) we get

${\text{P(C) = }}\dfrac{{^5{C_1}}}{{^{12}{C_1}}}$

Now using the formula of$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, let’s calculate the value of $\dfrac{{^5{C_1}}}{{^{12}{C_1}}}$hence it will be

$ \Rightarrow \dfrac{{\dfrac{{5!}}{{1!\left( {5 - 1} \right)!}}}}{{\dfrac{{12!}}{{1!\left( {12 - 1} \right)!}}}} = \dfrac{{\dfrac{{5!}}{{4!}}}}{{\dfrac{{12!}}{{11!}}}}$………………………………… (12)

Now using $n! = n \times (n - 1) \times (n - 2) \times (n - 3)...........(n - r){\text{ r < n}}$equation (8) can be simplified as

$ \Rightarrow \dfrac{{\dfrac{{5 \times 4!}}{{4!}}}}{{\dfrac{{12 \times 11!}}{{11!}}}} = \dfrac{5}{{12}}$

Hence the probability of drawing a black ball from the bag containing 3 red, 5 black and 4 white balls is$\dfrac{5}{{12}}$.

$(d)$ Non red ball -

Let D be the event of drawing a non-red ball.

Now in total we have 9 non-red balls (5 black and 4 white), thus the number of ways of selecting one non-red ball out of 9 non-red ball is $^9{C_1}$

Thus favorable outcomes for event D is$^9{C_1}$……………………… (13)

Now there are in total 12 balls in the bag, so the ways of selecting one ball out of these 12 balls will be$^{12}{C_1}$.

Thus the total number of possible outcomes for the event D is $^{12}{C_1}$…………………. (14)

Now probability of an event A that is ${\text{P(D) = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcome}}}}$……………………… (15)

Thus substituting the values from equation (13) and equation (14) in equation (15) we get

${\text{P(D) = }}\dfrac{{^9{C_1}}}{{^{12}{C_1}}}$

Now using the formula of$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, let’s calculate the value of $\dfrac{{^9{C_1}}}{{^{12}{C_1}}}$hence it will be

$ \Rightarrow \dfrac{{\dfrac{{9!}}{{1!\left( {9 - 1} \right)!}}}}{{\dfrac{{12!}}{{1!\left( {12 - 1} \right)!}}}} = \dfrac{{\dfrac{{9!}}{{8!}}}}{{\dfrac{{12!}}{{11!}}}}$………………………………… (12)

Now using $n! = n \times (n - 1) \times (n - 2) \times (n - 3)...........(n - r){\text{ r < n}}$equation (8) can be simplified as

$ \Rightarrow \dfrac{{\dfrac{{9 \times 8!}}{{8!}}}}{{\dfrac{{12 \times 11!}}{{11!}}}} = \dfrac{9}{{12}} = \dfrac{3}{4}$

Hence the probability of drawing a non-red ball from the bag containing 3 red, 5 black and 4 white balls is$\dfrac{3}{4}$.

Note: Whenever we face such types of problems the key concept that we need to apply is simply to find the number of favorable outcomes to that particular event and then the total possible outcome for that particular event. This will help get the right answer to the problem.

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