3 white ball,4 green ball,5 red ball.Probablity to drawn three balls of different colours
Answers
Even though there are 3 white balls, 4 green balls and 5 red balls in total, the total no. of outcomes is not 3 + 4 + 5 = 12 !
Three balls are drawn here simultaneously. When three balls are drawn, possibly we get either the following:
1. Three white balls.
2. Three green balls.
3. Three red balls.
4. One white ball and two green balls.
5. Two white balls and one green ball.
6. One green ball and two red balls.
7. Two green balls and one red ball.
8. One red ball and two white balls.
9. Two red balls and one white ball.
10. One white ball, one green ball and one red ball.
These are the total possible outcomes. Hence the total no. of outcomes is 10.
If white balls, green balls and red balls are indicated as W, G and R respectively, then the possible outcomes will be,
1. (W, W, W)
2. (W, W, G)
3. (W, W, R)
4. (G, G, G)
5. (G, G, R)
6. (G, G, W)
7. (R, R, R)
8. (R, R, W)
9. (R, R, G)
10. (W, R, G)
Okay, among these 10 total possibilities, there is only 1 outcome in which three different balls are drawn simultaneously.
(W, R, G)
Hence the no. of favourable outcomes is 1.
Thus the probability is 1/10.
Answer:
3/11
Step-by-step explanation:
white balls =3
green balls = 4
red balls = 5
total balls = 3+4+5 = 12
number of balls taken = 3
Total number of ways of selection= n[S] = 12C3 = 12x11x10 /1x2x3 =220
NUMBER OF WAYS
TO TAKE ONE FROM EACH COLOUR =n[A] = 3C1x4C1x5C1
= 3x4x5
=60
.Probability to drawn three balls of different colours = n[A]/n[S]
= 60/220
= 3/11