Question

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?

Answer 1

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

$P(AY n BG) = 0.2 * 0.2 = 0.04 -------(1)\\P(AG n BY) =0.1 * 0.14 =0.014 --------(2)$

The draws can happen in either of the 2 ways in (1) and (2) above

therefore total probability E is given as

E =$P( AY n BG) u P(AG n BY)\\=0.04 + 0.014 =0.O54$

For the yellow one to be from 1994, it implies that the event to be chosen is

$P(AYnBG) = 0.2*0.2$

Since the total probability is given as E=0.054

then $P((AYnBG) /E) =\frac{0.04}{0.054} = 0.07407$

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