Question

myopenmath.comAccording to Masterfoods, the company that manufactures M M's, 12% of peanut M&M's arebrown, 15% are yellow, 12% are red, 23% are blue, 23% are or nge and 15% are green.(Round your answers to 4 decimal places where possible)a. Compute the probability that a randomly selected peanut M M is not red.b. Compute the probability that a randomly selected peanut M8 is orange or blue.c Compute the probability that three randomly selected peanut! M's are all brown.d. If you randomly select five peanut M&M's, compute that probably that none of them are yellowe. If you randomly select five peanut M&M's, compute that probability that at least one of them isyellow.

Answer 1

You know that the colors of M&M's have the following distribution:

• 12% are brown (Br)

,

• 15% are yellow (Y)

,

• 12% are red (R)

,

• 23% are blue (Bl)

,

• 23% are orange (O)

,

• 15% are green (G)

Using these proportions you have to calculate several probabilities.

If oneM&M is chosen at random you have to determine the probability that:

a) "The M&M is not red"

This event includes all other possible colors, brown, yellow, blue, orange, and green. Since all other colors are included, you can conclude that the event "not red" is complementary to the event "red"

As you know, the sum of the probabilities of two complementary events is equal to one so that:

$P(R)+P(R^c)=1$

Where

P(R) is the probability of choosing a red M&M

P(R^c) is the probability of the complement, i.e. of not choosing a red M&M

The proportion or probability of choosing a red M&M is 12% or 0.12, then the complement can be calculated as:

$\begin{gathered} P(R^c)=1-P(R) \\ P(R^c)=1-0.12 \\ P(R^c)=0.88 \end{gathered}$

So, the probability that the M&M chosen is not red is 0.88.

b) "The M&M is orange or blue"

In this case, you have to calculate the probability of the union of both events, the M&M chosen is either orange or blue. The word "or" is the one that indicates that you have to work with the union of both events.

You can express this probability as follows:

$P(O\cup Bl)$

In this case, the events "orange" and "blue" are mutually exclusive, which means that they cannot occur at the same time. The union of two mutually exclusive events is equal to the sum of the probability of both events so that:

$P(O\cup Bl)=P(O)+P(Bl)$

The probabilities of both events are known, then the probability of choosing either an orange M&M or a blue M&M is:

$\begin{gathered} P(O\cup Bl)=0.23+0.23 \\ P(O\cup Bl)=0.46 \end{gathered}$

So, the probability that the M&M is either orange or blue is 0.46.

c) Three M&M's are selected at random, you have to determine the probability that all of them are brown.

That is "brown the first" and "brown the second" and "brown the third". The word "and" indicates that you have to work with the intersection of all events,

In this case, you have to calculate the probability of the intersection of three events:

$P(Br_1\cap Br_2\cap Br_3)$

Which is equal to the product of the probability of each event:

$P(Br_1\cap Br_2\cap Br_3)=P(Br_1)\cdot P(Br_2)\cdot P(Br_3)$

We know that the proportion of brown M&M's is 12%, which means that the probability of choosing a brown M&M is 0.12. Then:

$\begin{gathered} P(Br_1\cap Br_2\cap Br_3)=0.12\cdot0.12\cdot0.12 \\ P(Br_1\cap Br_2\cap Br_3)=0.12^3 \\ P(Br_1\cap Br_2\cap Br_3)=0.001728\approx0.017 \end{gathered}$

The subindices indicate the order in which each M&M is selected.

So, the probability that all three M&M's chosen are brown is 0.017.

d) If five M&M's are randomly selected, you have to determine the probability that none of them are yellow. That is that "the first one is not yellow" and "the second one is not yellow" and " the third one is not yellow" and "the fourth one is not yellow" and "the fifth one is not yellow"

$P(Y^c_1\cap Y^c_2\cap Y^c_3\cap Y^c_4\cap Y^c_5)=P(Y^c_1)\cdot P(Y^c_2)\cdot P(Y^c_3)\cdot P(Y^c_4)\cdot P(Y^c_5)$

The subindices indicate the order in which each M&M is selected.

The event "the M&M is not yellow" is complementary to the event "the M&M is yellow", first, calculate the probability of the complement:

$\begin{gathered} P(Y^c)=1-P(Y) \\ P(Y^c)=1-0.15 \\ P(Y^c)=0.85 \end{gathered}$

Then calculate the probability:

$\begin{gathered} P(Y^c_1\cap Y^c_2\cap Y^c_3\cap Y^c_4\cap Y^c_5)=0.85\cdot0.85\cdot0.85\cdot0.85\cdot0.85 \\ P(Y^c_1\cap Y^c_2\cap Y^c_3\cap Y^c_4\cap Y^c_5)=0.85^5 \\ P(Y^c_1\cap Y^c_2\cap Y^c_3\cap Y^c_4\cap Y^c_5)=0.4437 \end{gathered}$

So, the probability that none of five M&M's selected at random are not yellow is 0.4437

e) Out of five M&M's selected at random you have to compute the probability that at least one of them is yellow.

There are several situations where this description is true:

"all of them are yellow" or "one is yellow" or "two are yellow" or "three are yellow" or "four are yellow"

Symbolized as:

$\begin{gathered} P(Y_1\cap Y_2\cap Y_3\cap Y_4\cap Y_5)+5\lbrack P(Y_1\cap Y^c_2\cap Y_{^c3}\cap Y^c_4\cap Y^c_5)\rbrack+6\lbrack P(Y_1\cap Y_2\cap Y^c_3\cap Y^c_4\cap Y^c_5)\rbrack+6P(Y_1\cap Y_2\cap Y_3\cap Y^c_4\cap Y^c_5)+5\lbrack P(Y_1\cap Y_2\cap Y_3\cap Y_4\cap Y^c_5)\rbrack \\ =0.15^5+5(0.15\cdot0.85^4)+6(0.15^2\cdot0.85^3)+6(0.15^3\cdot0.85^2)+5(0.15^4\cdot0.85) \\ =0.49126\approx0.4913 \end{gathered}$

So, the probability of choosing at least one yellow M&M is 0.4913