Candy Probability

[Messagehelp]

The Masterfoods Company says that before the introduction of purple, yellow candies made up 20% of their plain M&M's, red another 20%, and orange, blue, and green each made up 10%. The rest were brown.

If you pick 3 M&M's in a row, what is the probability that
1)they are all brown?
2)the third one is the first that is red?
3) none are yellow?
4) at least one is green?

The part I am confused on is that I always thought you found probability by using the total number of possible occurances as the denominator and in this case I was never given the actual total number of M&M's just percentages. For example "they are all brown" at least if I was told there were 100 M&M's i could figure out the first probability would be 30/100, then 29/99, 28/98. For all I know it could have been out of 10 M&M's giving me 3/10, 2/9, and 1/8. But thats not what I was given, just percentages. So, I believe these questions could not be answered...which would seem odd for a problem. So am I lost? or was not enough information given.

 

[soroban]

Hello, Messagehelp!

These are independent events.
The probability of a yellow M&M is always 20% ... every time we select a candy.
The problem is actually much simpler.

The Masterfoods Company says: yellow made up 20% of their plain M&M's, red another 20%.
Orange, blue, and green each made up 10%. The rest were brown.

$\displaystyle \text{We are given: }\;\begin{array}{cccccccc} P(Y) &=& 0.2 & & P(\sim\! Y) &=& 0.8 \\ P(R) &=& 0.2 && P(\sim\! R) &=& 0.8 \\ P(O) &=& 0.1 \\ P(Bl) &=& 0.1 \\ P(G) &=& 0.1 && P(\sim\! G) &=& 0.9 \\ P(Br) &=& 0.3 \end{array}$

If you pick 3 M&M's in a row, what is the probability that:

1) they are all brown?

$\displaystyle \text{We want: }\: \text{Brown, Brown, Brown}$

$\displaystyle P(Br \wedge Br\wedge Br) \;=\;(0.3)(0.3)(0.3) \;=\;\boxed{0.027}$

2) the third one is the first that is red?

$\displaystyle \text{We want: }\:\sim\!R,\:\sim\!R,\:R$

$\displaystyle P(\sim\!R \:\wedge \sim\!R \wedge R) \;=\;(0.8)(0.8)(9.2) \;=\;\boxed{0.128}$

3) none are yellow?

$\displaystyle \text{We want: }\sim\!Y,\:\sim\!Y,\:\sim\!Y$

$\displaystyle P(\sim\!Y \wedge \sim\!Y \wedge \sim\!Y) \;=\;(0.8)(0.8)(0.8) \;=\;\boxed{0.512}$

4) at least one is green?

The opposite of "at least one green" is "no green."

$\displaystyle \text{We want: }\sim\!G,\:\sim\!G,\:\sim\!G$

$\displaystyle P(\sim\!G\: \wedge\sim\!G\:\wedge\sim\!G) \;=\;(0.9)(0.9)(0.9) \;=\;0.729$


\(\displaystyle \text{Since }P(\text{no G}) \;=\;0.729\,\text{then: }\(\text{at least one G}) \;=\;1 - 0.729 \;=\;\boxed{0.271}\)