Question

A bag contains 5 white, 6 red, 2 green and 2 black balls. One ball is selected at random from the bag. Find the probability that the selected ball is- i. white ii. non-white iii. white or green iv. black or red

Answer 1

1-5/15=1/3
2-10/15=2/3
3-7/15
4-8/15..........

Answer 2

Hey There!!

Here,

Number of white balls = 5
Number of red balls = 6
Number of green balls = 2
Number of black balls = 2


So, Total Number of balls = 5+6+2+2 = 15  balls



Now, 

1) Let Event A = Selected Ball is White

No. of outcomes favourable to A = 5 [as there are 5 white balls]


So,

$P(A) = \frac{5}{15} \\ \\ \implies \boxed{P(A)=\frac{1}{3}}$


2) Let Event B = Selected Ball is Non-White

No. of outcomes favourable to B = 10 [as there are 10 non-white balls]


So,

$P(B) = \frac{10}{15} \\ \\ \implies \boxed{P(B)=\frac{2}{3}}$


3) Let Event C = Selected Ball is White or Green


No. of outcomes favourable to C = 7 [5 white + 2 green balls]


So,
$\boxed{P(C)=\frac{7}{15}}$


4) Let Event D = Selected Ball is Blackor Red


No. of outcomes favourable to D = 8 [6 red + 2 black balls]


So,

$\boxed{P(D)=\frac{8}{15}}$


Hope this helps
Purva
Brainly Community