Question

a bag contains 3 white 5 black and 2 red balls
find the probability of getting :-
1) a white ball
2) not a red ball
3) not a black ball

Answer 1

hi friend

here is your answer looking for

total number of balls = 3+5+2
= 10 balls

so , the probability of given balls will be

$\frac{number \: of \: given \: balls \: }{number \: of \: total \: balls}$

1) probability of white ball
$= \frac{3}{10}$

2) probability of non red balls

$\frac{3 + 5}{10} \\ \\ = \frac{8}{10} \\ \\ = \frac{4}{5}$
3) probability of non black balls

$= \frac{3 + 2}{10} \\ \\ = \frac{5}{10} \\ \\ = \frac{1}{2}$
hope you are satisfied with my answer

Answer 2

Hello friends ⏬⏬

______________________________

Given :

white ball = 3

black ball = 5

red ball = 2

We have to find the probability getting:

1.) a white ball => ?

2.) not a red ball => ?

3.) not a black ball => ?

Let solve,

: Total no. of possible outcome = 3+5+2

= 10

1.) Let E be the event of getting white ball

The no. of outcomes in favorable to event E = 3

hence,

P(E) = P(White ball)
$= \frac{no. \: of \: outcomefavorble \: to \: event \: E}{no. \: of \: possible \: possible \: outcome}$

$= \frac{3}{10} \\ \\ P(white \: ball) = \frac{3}{10}$

2.) Let F is the event of getting not a red ball.

: The no. of outcomes favorable to event F
= 10 - 2

= 8

hence,

P(F)= P(not a red ball)
$= \frac{no. \: of \: outcomes \: favorable \: to \: event \: F}{no. \: of \: possible \: outcomes}$
$= \frac{8}{10} \\ \\ = \frac{4}{5} \\ \\ P(not \: a \: red \: ball) = \frac{4}{5}$

3.) Let B is the event of getting not a black ball.

The no. of outcomes favorable to event B .
= 10 - 5

= 5

hence,

P(B) = P(not a black ball)
$= \frac{no. \: of \: outcomes \: favorable \: to \: evenet \: B}{no. \: of \: possible \: outcomes} \\ \\ = \frac{5}{10} \\ \\ = \frac{1}{2} \\ \\ P(not \: a \: black \: ball) = \frac{1}{2}$

Hope Helpful..

:)