Question

There are 12 balls in a bag, 8 red and 4 green. Three
balls are drawn successively without replacement. What
is the probability that they are alternatively of the same
colour ?​

Answer 1

Answer:

No

Step-by-step explanation:

He randomly choose therefore they are not alternatively of the same colour

Answer 2

Answer:

probability of getting alternate colors when three balls are drawn without replacement  = $\frac{8}{33}$

Step-by-step explanation:

given, total number of balls = 12

number of red balls = 8

number of green balls = 4

to find : probability of getting alternate colors when three balls are drawn without replacement

there are 2 cases

case 1 : first ball red, second green and third red

P[first ball red] = $\frac{8}{12}$ = $\frac{2}{3}$

P[second green] = $\frac{4}{11}$  (1 ball is taken first, so there are 11 balls left)

P[third red] = $\frac{7}{10}$

∴ P[first ball red, second green and third red] = $\frac{2}{3} * \frac{4}{11} *\frac{7}{10}$ = $\frac{56}{330}$

         = $\frac{28}{165}$

case 2; first ball green, second red and third green

P[first ball green] = $\frac{4}{12} = \frac{1}{3}$

P[second red] = $\frac{8}{11}$

P[third green] = $\frac{3}{10}$

∴P[first ball green, second red and third green] = $\frac{1}{3} *\frac{8}{11} *\frac{3}{10}$ = $\frac{24}{330}$

            =$\frac{12}{165}$

required probability = P[first ball red, second green and third red] + P[first ball red, second green and third red]

required probability = $\frac{28}{165} + \frac{12}{165}$ = $\frac{40}{165}$

                  = $\frac{8}{33}$

therefore, probability of getting alternate colors when three balls are drawn without replacement  = $\frac{8}{33}$