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:

Three balls are drawn successively without replacement. What is the probability that they are alternatively of the same colour ...

Answer 2

Answer:

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

Step-by-step explanation:

given, total number of balls (green+ red) = 12

number of red balls = 8

number of green balls = 4

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

the two cases are as follows:

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

Probability of first ball red = $\frac{8}{12}$  = $\frac{1}{3}$

Probability of second ball green = $\frac{4}{11}$  (1 ball is drawn first, so there are 11 balls left)

Probability of third ball red = $\frac{7}{10}$

∴ Probability of first ball red, second ball green and third ball red :

P[RGR] = $\frac{2}{3}$ × $\frac{4}{11}$ × $\frac{7}{10}$ =  $\frac{56}{330}$

        = $\frac{28}{165}$

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

Probability of first ball green = $\frac{4}{12}$ = $\frac{1}{3}$

Probability of second ball red = $\frac{8}{11}$

Probability of third ball green = $\frac{3}{10}$

∴Probability of first ball green, second ball red and third ball green:

P[GRG] = $\frac{1}{3}$  ×  $\frac{8}{11}$ ×   $\frac{3}{10}$ = $\frac{24}{330}$

           = $\frac{12}{165}$

required probability = P[GRG] + P[RGR]

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

                 = $\frac{8}{33}$

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