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