Chapter 5, Problem 60CE

M&Ms are blended in a ratio of 13 percent brown. 14 percent yellow, 13 percent red, 24 percent blue, 20 percent orange, and 16 percent green. Suppose you choose a sample of two M&Ms at random from a large bag. (a) Show the sample space. (b) What is the probability that both are brown? (c) Both blue? (d) Both green? (e) Find the probability of one brown and one green M&M. (f) Actually take 100 samples of two M&Ms (with replacement) and record the frequency of each outcome listed in (b) and (c) above. How close did your empirical results come to your predictions? (g) Which definition of probability applies in this situation?

To determine

Show the sample space.

Explanation of Solution

The given information is that the M&Ms were blended in the ratio of 13% brown, 14% yellow, 13% red, 24% blue, 20% orange and 16% green. Also, the selected random sample of M&M blended in the large bag is 2.

Sample space:

The set of all possible outcomes of the random experiment is said to be sample space of and it is denoted by S.

Consider, B represents brown, Y represents yellow, R represents red, BL represents blue, O represent orange and G represents green.

The all possible outcomes of the random experiment of any two of brown, yellow, red, blue, orange and green M&Ms is as follows.

$S=\left\{\begin{array}{l}\left(B,B\right),\left(B,Y\right),\left(B,R\right),\left(B,BL\right),\left(B,O\right),\left(B,G\right),\\ \left(Y,B\right),\left(Y,Y\right),\left(Y,R\right),\left(Y,BL\right),\left(Y,O\right),\left(Y,G\right),\\ \left(R,B\right),\left(R,Y\right),\left(R,R\right),\left(R,BL\right),\left(R,O\right),\left(R,G\right),\\ \left(BL,B\right),\left(BL,Y\right),\left(BL,R\right),\left(BL,BL\right),\left(BL,O\right),\left(BL,G\right),\\ \left(O,B\right),\left(O,Y\right),\left(O,R\right),\left(O,BL\right),\left(O,O\right),\left(O,G\right),\\ \left(G,B\right),\left(G,Y\right),\left(G,R\right),\left(G,BL\right),\left(G,O\right),\left(G,G\right)\end{array}\right\}$.

To determine

Find the probability that both are brown.

Answer to Problem 60CE

The probability that both are brown is 0.0169.

Explanation of Solution

Calculation:

From the given information, it can be observed that the probability of brown M&M is 0.13 $\left(=\frac{13}{100}\right)$.

The probability that both are brown is obtained as follows:

$\begin{array}{c}P\left(\text{Both are brown}\right)=P\left(B\right)\times P\left(B\right)\\ =0.13\times 0.13\\ =0.0169\end{array}$

Thus, the probability that both are brown is 0.0169.

To determine

Find the probability that both are blue.

Answer to Problem 60CE

The probability that both are blue is 0.0576.

Explanation of Solution

Calculation:

From the given information, it can be observed that the probability of blue M&M is 0.24 $\left(=\frac{24}{100}\right)$.

The probability that both are blue is obtained as follows:

$\begin{array}{c}P\left(\text{Both are blue}\right)=P\left(BL\right)\times P\left(BL\right)\\ =0.24\times 0.24\\ =0.0576\end{array}$

Thus, the probability that both are blue is 0.0576.

To determine

Find the probability that both are green.

Answer to Problem 60CE

The probability that both are green is 0.0256.

Explanation of Solution

Calculation:

From the given information, it can be observed that the probability of green M&M is 0.16 $\left(=\frac{16}{100}\right)$.

The probability that both are green is obtained as follows:

$\begin{array}{c}P\left(\text{Both are green}\right)=P\left(G\right)\times P\left(G\right)\\ =0.16\times 0.16\\ =0.0256\end{array}$

Thus, the probability that both are green is 0.0256.

To determine

Find the probability of one brown and one green M&M.

Answer to Problem 60CE

The probability of one brown and one green M&M is 0.0208.

Explanation of Solution

Calculation:

The probability of one brown and one green M&M is obtained as follows:

$\begin{array}{c}P\left(\text{One brown and one green}\right)=P\left(B\right)\times P\left(G\right)\\ =0.13\times 0.16\\ =0.0208\end{array}$

Thus, the probability of one brown and one green M&M is 0.0208.

To determine

Take 100 samples of two M&Ms (with replacement). Record the frequencies of each outcome listed in (b) and (c) and how close the empirical results come to the people prediction.

Explanation of Solution

Calculation:

Answers may vary. One of the possible answers is as follows:

All 100 samples of two M&Ms (with replacement) is,

$\left\{\begin{array}{l}\left(G,B\right),\left(R,Y\right),\left(Y,R\right),\left(BL,BL\right),\left(G,O\right),\left(Y,G\right),\left(R,Y\right),\left(Y,R\right),\left(O,O\right),\left(O,BL\right),\\ \left(B,BL\right),\left(B,G\right),\left(R,Y\right),\left(O,G\right),\left(G,Y\right),\left(Y,R\right),\left(R,B\right),\left(Y,Y\right),\left(O,BL\right),\left(BL,G\right),\\ \left(G,O\right),\left(O,O\right),\left(BL,R\right),\left(Y,BL\right),\left(R,G\right),\left(BL,BL\right),\left(BL,BL\right),\left(R,Y\right),\left(G,R\right),\left(O,G\right),\\ \left(Y,B\right),\left(R,O\right),\left(O,B\right),\left(R,R\right),\left(B,B\right),\left(B,Y\right),\left(BL,O\right),\left(Y,R\right),\left(R,G\right),\left(G,O\right),\\ \left(B,R\right),\left(R,R\right),\left(Y,BL\right),\left(BL,BL\right),\left(B,B\right),\left(BL,B\right),\left(BL,B\right),\left(O,G\right),\left(R,Y\right),\left(B,G\right),\\ \left(Y,B\right),\left(Y,Y\right),\left(Y,R\right),\left(Y,BL\right),\left(Y,O\right),\left(Y,G\right),\left(BL,B\right),\left(BL,Y\right),\left(BL,R\right),\left(BL,BL\right),\\ \left(B,B\right),\left(R,Y\right),\left(B,B\right),\left(BL,BL\right),\left(R,O\right),\left(G,R\right),\left(BL,O\right),\left(R,BL\right),\left(O,G\right),\left(G,G\right),\\ \left(Y,Y\right),\left(R,Y\right),\left(O,O\right),\left(R,Y\right),\left(R,R\right),\left(G,G\right),\left(BL,BL\right),\left(B,BL\right),\left(B,B\right),\left(Y,G\right),\\ \left(B,B\right),\left(B,G\right),\left(G,BL\right),\left(R,O\right),\left(O,O\right),\left(O,O\right),\left(R,Y\right),\left(Y,Y\right),\left(BL,BL\right),\left(R,B\right),\\ \left(BL,R\right),\left(G,O\right),\left(O,O\right),\left(G,BL\right),\left(B,B\right),\left(Y,R\right),\left(BL,BL\right),\left(O,G\right),\left(G,O\right),\left(B,BL\right)\end{array}\right\}$

The number of frequencies for both are brown is 7 and the number of frequencies for both are blue is 9 and a total number frequency is 100.

For (b) probability that both are brown:

The probability that both are brown is obtained as follows:

$\begin{array}{c}P\left(\text{Both are brown}\right)=\frac{N\left(\text{Number of frequencies for both are brown}\right)}{\text{Total number of frequencies}}\\ =\frac{7}{100}\\ =\frac{7}{100}\\ =0.07\end{array}$

Thus, the probability that both are brown is 0.07.

For (c) probability that both are blue:

The probability that both are blue is obtained as follows:

$\begin{array}{c}P\left(\text{Both are blue}\right)=\frac{N\left(\text{Number of frequencies for both are blue}\right)}{\text{Total number of frequencies}}\\ =\frac{9}{100}\\ =\frac{9}{100}\\ =0.09\end{array}$

Thus, the probability that both are blue is 0.09.

From part (b) and (c), the probability that both are brown is 0.0169 and the probability that both are blue is 0.0576.

From the results, it is observed that the empirical results are greater than the results of part (b) and (c). That is, $0.07>0.0169$ and $0.09>0.0576$.

To determine

Identify which definition of probability applies in the given situation.

Answer to Problem 60CE

The definition of probability applies in the given situation is empirical.

Explanation of Solution

Classical Probability:

If each outcome in a sample space is equally likely to occur then it is said to be classical probability.

Empirical Probability:

The observations which are found from probability experiment is termed as empirical probability.

Subjective Probability:

If the probabilities result from intuition, educated guesses and estimates, then the probability is said to be subjective probability.

Here, the situation is most likely based on an experiment. That is, the number of each outcome is noted and the corresponding probabilities are obtained.

Thus, the definition of probability applies in the given situation is empirical.