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?

Jul 14, 2016

$P \left(\text{RRG}\right) = \frac{72}{1331}$

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)

=$\frac{3}{11} \times \frac{3}{11} \times \frac{8}{11}$

= $\frac{72}{1331}$

Jul 14, 2016

Reqd. Prob.$= \frac{72}{1331.}$

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.$= P \left({R}_{1} \cap {R}_{2} \cap {G}_{3}\right)$
$= P \left({R}_{1}\right) \cdot P \left({R}_{2} / {R}_{1}\right) \cdot P \left({G}_{3} / \left({R}_{1} \cap {R}_{2}\right)\right) \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right)$

For $P \left({R}_{1}\right) : -$

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, $P \left({R}_{1}\right) = \frac{3}{11}$.......(2)

For $P \left({R}_{2} / {R}_{1}\right) : -$

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, $P \left({R}_{2} / {R}_{1}\right) = \frac{3}{11.} \ldots \ldots \ldots \left(3\right)$

Finally, on the same line of arguments, we have, $P \left({G}_{3} / \left({R}_{1} \cap {R}_{2}\right)\right) = \frac{8}{11.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(4\right)$

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

Reqd. Prob.$= \frac{3}{11} \cdot \frac{3}{11} \cdot \frac{8}{11} = \frac{72}{1331.}$

Hope, this will be helpful! Enjoy Maths.!