Math, asked by slnchary4610, 1 year ago

A box contains 5 red, 4 white and 3 green balls. In how many ways can 3 balls be drawn from the box, without replacement, so that at least 2 of them are green?

Answers

Answered by MavisRee
29

Answer:

The Answer is 23 / 110

Step-by-step explanation:

Given,

A box contains 5 red, 4 white and 3 green balls

Total number of balls = 5 + 4 + 3

Three balls randomly can be picked in  ¹²C₃ ways, that is,

220 ways

According to question,

Atleast 2 of the balls should be green :

CASE 1 :

2 Green balls, 1 Other ball

Number of ways 2 Green balls can be drawn out of total 3 green balls is :

³C₂ ways = 3 ways

1 other balls from the other left balls = ⁹C ₁ = 9 ways

So,

2 Green balls and 1 other ball, Number of ways = 3 * 9 = 45 ways

Case 2 :

All the 3 balls are Green

Number of ways = ³C₃ = 1 way

So, three balls drawn in (45 + 1) = 46 ways will have at least two green balls among the drawn ones

So, the probability of getting at least three Green balls among the three drawn =

( 46 / 220 ) = ( 23 / 110 )

Hence,

The Answer is 23 / 110


Answered by usmanfaskari12
24

Answer:

Step-by-step explanation:

A box contain 3 red, 4 white and 5 green balls. Three balls are drawn at random and replace one after the other. What is the probability that the first is green, the second is white and the third is green ( green, white, green)

Similar questions