There is a problem: A jar contains 6 red jelly beans, 6 green jelly beans, and 6 blue jelly beans. If we choose a jelly bean, then another jelly bean without putting the first one back in the jar, what is the probability that the first jelly bean will be green and the second will be green as well?
So we need to find P(2ndG|1stG), and somehow i solve it using intuition means that if P(G)=3/18=1/3 and P(another green) = 5/17, result will be 5/51, and this is right result, but i was wondering what is the right way to solve that problem with some mathematical rather than intuitive way.
I tried to use Bayes' theorem P(2ndG|1stG)=P(1stG|2ndG)*P(1stG)/P(2ndG), but i do not see any sense of solving it since i do not understand how to calculate P(1stG|2ndG) or P(2ndG)...
Please help me what i get wrong...
Since we have $6$ green beans and $18$ beans in total, probability that the first bean picked is green can be calculated as
$$P(G_1) = \frac{6}{18}$$
Now if the first bean picked was green, then we are left with $5$ green beans and $17$ beans in total, hence probability of picking second green bean is
$$P(G_2 \mid G_1) = \frac{5}{17}$$
And therefore the probability that both beans picked are green is calculated as
$$P(G_1 \cap G_2) = \frac{6}{18} \cdot \frac{5}{17}$$