The blue m& m was introduced in 1995. before then, the color mix in a bag of plain m& ms was (30% brown, 20% yellow, 20% red, 10% green, 10% orange, 10% tan). afterward it was (24% blue , 20% green, 16% orange, 14% yellow, 13% red, 13% brown). a friend of mine has two bags of m& ms, and he tells me that one is from 1994 and one from 1996. he won't tell me which is which, but he gives me one m& m from each bag. one is yellow and one is green. what is the probability that the yellow m& m came from the 1994 bag?

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

it should be a and d. lmk if i'm wrong xd

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step-by-step explanation: