yashsokhal23Box 1:
6 green +
4 redBox 2:
3 green +
7 redNow there are two routes here:
1) Green is drawn from box 1 and then Green is drawn from box 2.
2) Red is drawn from box 1 and then Green is drawn from box 2.
When solving this question one needs to take into account the probability of drawing Green or Red in the first draw
AND then the probability of drawing Green from the second draw in either case and then see what the probability is of either the 1st case
OR second case.
1) Probability that Green is first drawn and then Green is drawn: \(\frac{6}{10}*\frac{4}{11} = \frac{12}{55}\)
2) Probability that Red is first drawn and then Green is drawn: \(\frac{4}{10}*\frac{3}{11}=\frac{6}{55}\)
Now as there are two possible outcomes, it's an OR case, so we add the two probabilities together:
\(\frac{12}{55}+\frac{6}{55}= \frac{18}{55}\)
Answer B