Question

According 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 orange 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 orange,b. Compute the probability that a randomly selected peanut M&M is red or yellow,c. Compute the probability that two randomly selected peanut M&M's are both green,d. If you randomly select two peanut M&M's, compute that probability that neither of them areorangee. If you randomly select two peanut M&M's, compute that probability that at least one of them isorange

Answer 1

SOLUTION:

Step 1:

In this question, we are given the following:

Step 2:

We have the following probabilities based on their colors:

$\begin{gathered} 12\text{ \% ---- Brown} \\ 15\text{ \% -----Yellow} \\ 12\text{ \% -----Red} \\ 23\text{ \% -----Blue} \\ 23\text{ \% -----Orange} \\ 15\text{ \% -----Green} \\ \text{Total = 100 \%} \end{gathered}$

a) Compute the probability that the randomly selected peanut M&M is not

orange:

$\begin{gathered} \text{Probability ( the randomly selected peanut M\& M is not orange)} \\ =\text{ 100\% - 23 \% = 77\% = 0.77} \end{gathered}$

Step 3:

b) Compute the probability that the randomly selected peanut M&M is red or yellow:

$\begin{gathered} \text{Probability ( the randomly selected peanut M\&M is red or yellow)} \\ =\text{ 12 \% + 15\% = 27 \% = 0.27} \end{gathered}$

Step 4:

c) Compute the probability that the randomly selected peanut M&M are both green

$\begin{gathered} \text{Probability( the randomly selected peanut M \& M are both gr}een\text{ )} \\ =\text{ 15 \% x 15 \% } \\ =\text{ 0. 15 x 0. 15} \\ =\text{ 0.0225 ( 4 decimal places)} \end{gathered}$

Step 5:

d) If you randomly select two peanuts M&M's, compute the probability that neither of them are orange

$\begin{gathered} P\text{ ( neither of them are orange ) = ( 100 - 23 ) \% x ( 100 - 23 ) \%} \\ =\text{ 77\% x 77\%} \\ =\text{ 0.77 x 0.77} \\ =\text{ 0.5929 ( 4 decimal places)} \end{gathered}$

Step 6:

e) If you randomly select two peanut M&M's, compute that probability that at least is orange:

In this case, we would consider six different options here:

i ) Probability ( Brown & Orange )

$\begin{gathered} =12\text{ \% x 23 \%} \\ =\text{ 0. 12 X 0. 23 } \\ =\text{ 0.0276} \end{gathered}$

ii ) Probability ( Yellow & Orange):

$\begin{gathered} =15\text{ \% x 23 \%} \\ =\text{ 0. 15 x 0. 23 } \\ =\text{ }0.0345 \end{gathered}$

iii )Probability ( Red & Orange):

$\begin{gathered} =12\text{ \% x 23\%} \\ =\text{ 0. 12 x 0. 23 } \\ =\text{ 0.0276} \end{gathered}$

iv) Probability ( Blue & Orange ):

$\begin{gathered} =\text{ 23 \% x 23 \% } \\ =\text{ 0. 23 x 0. 23} \\ =\text{ 0.0529} \end{gathered}$

v) Probability ( Orange & Orange):

$\begin{gathered} =23\text{ \% x 23\% } \\ =\text{ 0.23 x 0. 23} \\ =\text{ 0.0529} \end{gathered}$

vi ) Probability ( Green & Orange):

$\begin{gathered} =15\text{ \% x 23\%} \\ =\text{ 0. 15 x 0. 23} \\ =\text{ 0.0345} \end{gathered}$

Now, we need to sum up the individual probabilities, we have that:

$\begin{gathered} =\text{ 0.0276 + 0.0345 + 0.0276 + 0.0529 + 0.0529 + 0.0345} \\ =\text{ 0.23}00\text{ ( 4 decimal places)} \end{gathered}$