There are 3 red and 8 green balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probability of choosing 2 red balls and then 1 green ball?

2 Answers
Jul 14, 2016

#P("RRG") = 72/1331#

Explanation:

The fact that the ball is replaced each time, means that the probabilities stay the same each time a ball is chosen.

P(red, red, green) = P(red) x P(red) x P(green)

=# 3/11 xx 3/11 xx 8/11#

= #72/1331#

Jul 14, 2016

Reqd. Prob.#=72/1331.#

Explanation:

Let #R_1#= the event that a Red Ball is chosen in the First Trial

#R_2#=the event that a Red Ball is chosen in the Second Trial

#G_3#=the event that a Green Ball is chosen in the Third Trial

:. Reqd. Prob.#=P(R_1nnR_2nnG_3)#
#=P(R_1)*P(R_2/R_1)*P(G_3/(R_1 nnR_2))..................(1)#

For #P(R_1):-#

There are 3 Red + 8 Green = 11 balls in the bag, out of which, 1 ball can be chosen in 11 ways. This is total no. of outcomes.

Out of 3 Red balls, 1 Red ball can be chosen in 3 ways. This is no. of outcomes favourable to #R_1#. Hence, #P(R_1)=3/11#.......(2)

For #P(R_2/R_1):-#

This is the Conditional Prob. of occurrence of #R_2#
, knowing that #R_1# has already occurred. Recall that the Red ball chosen in R_1 has to be replaced back in the bag before a Red ball for R_2 is to be chosen. In other words, this means that the situation remains same as it was at the time of #R_1#. Clearly, #P(R_2/R_1)=3/11..........(3)#

Finally, on the same line of arguments, we have, #P(G_3/(R_1 nnR_2))=8/11.......................(4)#

From #(1),(2),(3),&(4),#

Reqd. Prob.#=3/11*3/11*8/11=72/1331.#

Hope, this will be helpful! Enjoy Maths.!