Question

A box contains 6 black and 4 white balls. If a ball is taken from it, what is the probalityof the being white? And what is the probality of it being black?

Answer 1

Answer:

white- 2/5

black - 3/5

Step-by-step explanation:

p(white)=4/10=2/5

p(black)=6/10=3/5

Answer 2

Step-by-step explanation:

${\Large{\mathrm{\underline{\orange{Question :}}}}}$

A box contains 6 black and 4 white balls. If a ball is taken from it, what is the probalityof the being white ? And what is the probality of it being black ?

$\\ {\Large{\mathrm{\underline{\orange{Solution :}}}}}$

Given :

• Number of black balls = 6

• Number of white balls = 4

• Total number of balls = 6 + 4 = 10 balls

One ball is taken from the box

To find :

• The probability of it being white
• The probability of it being black

Formula being used :

$\\ {\bold{\boxed{\mathrm{Probability \: = \: \frac{number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes}}}}}$

Solution :

Probability of taking white ball :

${\bold{\sf{⟼ \: \: \: \: \: P(E) \: = \: \frac{number \: of \: white \: balls}{Total \: number \: of \: balls}}}}$

${\bold{\sf{⟼ \: \: \: \: \: P(E) \: = \: \frac{4}{10}}}}$

$⟼ \: \: \: \: \: {\bold{\boxed{\sf{\pink{P(E) \: = \: \frac{2}{5}}}}}}$

Probability of taking black ball :

${\bold{\sf{⟼ \: \: \: \: \: P(E) \: = \: \frac{number \: of \: black \: balls}{Total \: number \: of \: balls}}}}$

${\bold{\sf{⟼ \: \: \: \: \: P(E) \: = \: \frac{6}{10}}}}$

$⟼ \: \: \: \: {\bold{\boxed{\sf{\pink{P(E) \: = \: \frac{3}{5}}}}}}$