Question

a bag contains white black and red balls only a ball is drawn at random from the bag the possibility of getting a white ball is 3 by 10 and that of a black ball is 2 by 5 find the possibility of getting a red ball if the bag contains 20 black balls then find the total number of the balls in the bag

Answer 1

white ball (unknown)
black ball 20
red ball (unknown)
ATQ
probability of getting a black ball = 2/5
20/total no of balls= 2/5
total no of balls =20*5/2
total no of balls = 50

ATQ
total no of white balls/total no of balls =3/10
total no of white balls = 3*50/10
total no of white balls = 15

probability of getting a red ball =
total no of red balls/total no of balls
50-(15+20) /50
15/50
3/10(ans)

Answer 2

$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\underline{\Huge{\textbf{Step-1:}}}$
$\sf\textsf{Note the given data from the Question}$

Given,
$\sf\textsf{Bag contains White, Black and Red balls.}$

No. of Black balls n(B) = 20

$\textbf{Experiment:\: \textsf{A ball is drawn randomly from the bag}}$

Let W be the Event of getting White ball

Let B be the Event of getting Black ball

Let R be the Event of getting Red ball

$P(W) = \dfrac{3}{10}$ ————[1]

$P(B) = \dfrac{2}{5}$ ————[2]

$P(R) = ?$



$\underline{\Huge{\textbf{Step-2:}}}$
$\sf\textsf{Assume two variables for number of White and Red balls in the bag}$

Let number of White balls in the bag be a

Let number of Red balls in the bag be b



$\underline{\Huge{\textbf{Step-3:}}}$
$\sf\textsf{Find the Total Outcomes when a ball is drawn from the Bag}$

Let S be the Sample space.
S consists of White, Black and Red balls.
S = W+B+R

Total Outcomes when a ball is drawn randomly from a bag = $^{(Total\: No.\: of\: Balls\: in\: Bag)}C_{1}$

$\implies n(S) = ^{a + b + 20}C_{1}$

$\implies n(S) = a+b+20$

We have,
$\bigstar\: \: \boxed{\displaystyle\Large{\mathbf{Probability\: of\: Occurrence\: of\: Event= \dfrac{No.\: of\: Favorable\: Outcomes\: of\: Event}{Total\: No.\: of\: Outcomes\: in\: Experiment}}}}$



$\underline{\Huge{\textbf{Step-4:}}}$
$\sf\textsf{Find the Probability of Occurrence of Event W}$

No. of Favorable Outcomes for Occurrence of Event W n(W) = No. of White Balls

$\implies n(W) = a$

$P(W) = \dfrac{n(W)}{n(S)}$

$\therefore$
$P(W) = \dfrac{a}{a+b+20}$ ————[3]



$\underline{\Huge{\textbf{Step-5:}}}$
$\sf\textsf{Find the Probability of Occurrence of Event B}$

No. of Favorable Outcomes for Occurrence of Event B n(B) = No. of Black Balls

$\implies n(B) = 20$

$P(B) = \dfrac{n(B)}{n(S)}$

$\therefore$
$P(W) = \dfrac{b}{a+b+20}$ ————[4]



$\underline{\Huge{\textbf{Step-6:}}}$
$\sf\textsf{Equate [1]&[3] and form an Equation in two variable}$

$\implies \dfrac{a}{a+b+20} = \dfrac{3}{10}$

$\implies 10a = 3(a+b+20)$

$\implies 10a = 3a+3b+60$

$\implies 7a-3b =60$ ————[5]



$\underline{\Huge{\textbf{Step-7:}}}$
$\sf\textsf{Equate [2]&[4] and form another Equation in two variable}$

$\implies \dfrac{20}{a+b+20} = \dfrac{2}{5}$

$\implies 100 = 2(a+b+20)$

$\implies 100 = 2a+2b+40$

$\implies 2a+2b =60$

$\implies a+b =30$ ————[6]

$\underline{\Huge{\textbf{Step-8:}}}$
$\sf\textsf{Solve [5]&[6] to find the values of a and b}$

Multiply [6] with 7 on both sides

$\implies 7a+7b = 210$ ————[7]

Do [7] - [6]

$\begin{array}{cccccccc}&7a&+&&7b&=&&210\\&7a&-&&3b&=&&60 \\-&&&+&&&-&\\&&&&10b&=&&150\end{array}$

$\implies b = 15$

Substitute value of in [6]

$\implies a+15 =30$

$\implies a = 15$

$\therefore$
No. of White balls in the Bag, a= 15
No. of Red Balls in the Bag, b = 15

$\therefore \: n(S)= 15+15+20= 50$

No. of Favorable Outcomes for Occurrence of Event R n(R) = No. of Red Balls

$\implies n(R) = b = 15$

$P(R) = \dfrac{n(R)}{n(S)}$

$\implies P(R) = \dfrac{15}{50}=\dfrac{3}{10}$



$\blacksquare\: \:\textsf{Probability of getting a Red ball = \underline{\Large{\textbf{ \dfrac{3}{10} }}}}$