Question

1. A bag contains 4 red, 6 yellow and 5
green balls. 3 balls are drawn randomly.
What is the probability that the balls drawn
contain balls of different colours? *
(2 Points)
​

Answer 1

Answer:

the probability that the ball drawn contain balls of different colouts is 1/11

Answer 2

Answer:

probability of drawing 3 balls of different colors = 0.26

Step-by-step explanation:

given, there are 4 red, 6 yellow and 5 green balls

total 15 balls

$nC_{r} = \frac{n!}{r!(n-r)!}$

number of ways of drawing 3 balls from 15 balls = $15C_{3}$

            = $\frac{15!}{3!(15-3)!}$

           = $\frac{15*14*13*12!}{3*2*1*12!}$

            = 455

number of ways of selecting 1 red ball from 4 = $4C_{1}$= 4

number of ways of selecting 1 yellow from 6 = $6C_{1}$= 6

number of ways of selecting 1 green from 5 = $5C_{1}$= 5

number of ways of selecting 3 balls of different colors

            = 4×6×5  = 120

probability of drawing 3 balls of different colors  = $\frac{4C_{1}*5C_{1}*6C_{1} }{15C_{3} }$

         = $\frac{120}{455}$ = $\frac{24}{91}$

        = 0.26