A box contain 12 balls in which 4 are white 3 are blue and 5 are red.3 balls are drawn at random from the box. find the chance that all three are selected

3/11

Step-by-step explanation:

From the above question, we have the following information

Total number of balls = 12

Number of white balls = 4

Number of blue balls = 3

Number of red balls = 5

We solve this question using combination formula

C(n, r) = nCr = n!/r!(n - r)!

We are told that 3 balls are drawn out at random.

The chance/probability of drawing out 3 balls = 12C3 = 12!/3! × (12 - 3)! = 12!/3! × 9!

= 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1/(3 × 2 × 1) × (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)

= 220 ways

The chance of selecting 3 balls at random = 220

To find the chance that all the three balls are selected,

= [Chance of selecting (white ball) × Chance of selecting(blue ball) × Chance of selecting(red balls)]/ The chance/probability of drawing out 3 balls

Chance of selecting (white ball)= 4C1

Chance of selecting(blue ball) = 3C1 Chance of selecting(red balls) = 5C1

Hence,

= [4C1 × 3C1 × 5C1]/ 220

= 60/220

= 6/22

= 3/11

The chance that all three are selected is = 3/11

yes the total will be same because in both of them the expression.is rearranged but their sign remains the same hence it hss just been shuffled