Question

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?

Answer 1

`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`

Answer 2

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.!