Ron has a bag containing 3 green pears and 4 red pears he randomly selects a lead then randomly selects another pear without replacement which three diagram shows the correct probabilities for this situation

First pear            Second pear      probability

green (3/7)green (2/6)         (3/7)*(2/6) = 1/7
                  red (4/6)             (3/7) *(4/6) = 2/7

red (4/7)green (3/6)        (4/7)*(3/6) = 2/7
             red (3/6)             (4/7)*(3/6) = 2/7
                                                    
                                                     1/7 + 2/7 + 2/7 + 2/7 = 1

The probailities that two are green are: 1/7

The probabilities that one is green and the other is red are:2 /7 + 2/7 = 4/7

The probabilities that two are red are: 2/7

The correct answer would be quadrent 4