Help me solve this problem
2 answers:
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Answer:
The probability that the yellow M&M came from the 1994 bag is 0.07407 or 7.407%
Step-by-step explanation:
Given
Before 1995
(Br) Brown = 30%
(Y) Yellow = 20% =0.2
(R) Red = 20%
(G) Green =10% =0.1
(O) Orange = 10%
(T) Tan = 10%
After 1995
(Br) Brown = 13%
(Y) Yellow = 14% =0.14
(R) Red = 13%
(G) Green = 20% = 0.2
(O) Orange = 16%
(Bl) Blue = 24%
Since there are two bags, let A be the bag from 1994, and B be the bag from 1996
Then let AY imply we drew a yellow M&M from the 1994 bag
AG implies we drew a green M&M from the 1994 bag
BY implies imply we drew a yellow M&M from the 1996 bag
BG implies we drew a green M&M from the 1996 bag
P(AY) =0.2
P (BY) = 0.14
P(AG) =0.1
P(BG) =0.2
Since the draws from the 1994 and 1996 bag are independent,
therefore
The draws can happen in either of the 2 ways in (1) and (2) above
therefore total probability E is given as
E =
For the yellow one to be from 1994, it implies that the event to be chosen is
Since the total probability is given as E=0.054
then
Concluding statement: This is the condition for the Yellow one to come from 1994 and green from 1996 provided that they obey the condition from E
I hope this helps you
