Question 621751: Hi, I would like some help with this problem:
A bag of M&M's contains 20 pieces of candy.Ethan purchased a bag that contained seven orange pieces six green pieces, four yellow pieces, and three red pieces.He selects one piece of candy from the bag at random.Find the probability that he will select
a) a red piece of candy.
Would that be 3/20?
b) a green piece of candy.
Would it be 6/20?
c) a red or yellow piece of candy.
d) neither a red or yellow piece of candy
e) anything but an orangepiece of candy.
f) a purple piece of candy
Would that be 0/20?
Thank you for your help.
Found 2 solutions by solver91311, ewatrrr:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
a) yes, 3/20
b) yes, 6/20, but you should simplify to 3/10
c) There are 3 reds plus 4 yellows, so there are 7 possible successful outcomes
d) there are 7 orange pieces, so there are 20 minus 7 = 13 of everything else and getting one from the everything else group is classified as success.
e) yes, but express your answer as 0.
John
My calculator said it, I believe it, that settles it
Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website!
Hi,
20 pieces of candy : 7 orange, 6 green, 4 yellow and 3 red
one piece of candy selected from the bag at random.
P(red) = 3/20 Yes. good work
P(green) = 6/20 Yes. good work
P(red 'or' yellow) = 3/20  4/20 = 7/20
P(neither red or yellow) = 1- 7/20 = 13/20
P( not orange)= 1 - P(orange) = 1 - 4/20 = 16/20 = 4/5
P( Purple) = 0 :) Yes. good work
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