Question

in a bag there are 6 red balls, 2 yellow balls and 4 white balls. find the probability of
(i) getting a yellow balls
(ii) getting any color ball
(iii) getting a red balls
(iv) not getting a yellow balls
(v) getting an orange ball​

Answer 1

Answer:

(i)-4/12=1/3

(ii)3/12=1/4

(iii)6/12=1/2

(iv)-10/12=5/6

(v)-0 (impossible event orange is not in bag )

Step-by-step explanation:

please mark as brainlist

Answer 2

Given

6 red balls ,2 yellow balls ,4 white balls

To Find

(i) getting a yellow balls

(ii) getting any color ball

(iii) getting a red balls

(iv) not getting a yellow balls

(v) getting an orange ball

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

Probability is the no .of favourable outcomes upon total no.of outcomes

$\sf\boxed{\bold{Probability = \dfrac{favourable \: outcomes}{total \: no.of \: of \: outcomes} }}$

Total outcomes =6+2+4=12

Favourable outcomes=2

(I) P(getting a yellow balls)=2/12=1/6

(ii)Total outcomes =6+2+4=12

Favourable outcomes=3(any color may be its red, yellow &white)

P(getting any color balls)=3/12=1/4

(iii)Total outcomes =6+2+4=12

Favourable outcomes=6

P(getting a yellow balls)=6/12=1/2

(iv)

Total outcomes =6+2+4=12

Favourable outcomes=6+4=10

P(getting a yellow balls)=10/12=5/6

(v)Total outcomes =6+2+4=12

Favourable outcomes=0

P(getting a yellow balls)=0/12=0

As there is no orange balls so, 0