Chapter 6, Problem R6.5RE

(a)

To determine

Whether it is reasonable to use binomial distribution for calculating the probability involves X .

Answer to Problem R6.5RE

It is reasonable that X has a binomial distribution.

Explanation of Solution

Given information:

Number of trials, $n=8$

Probability of success, $p=20.5\%=0.205$

Four conditions of binomial setting are as follows:

• Binary (success/failure)
• Independent trials
• Fixed number of trials
• Probability of success (same for each trial)

Binary: Since the success results in orange candy and failure results in candy is not colored orange, the condition has been satisfied.

Independent trials: Since the random sample of 8 candies is less than 10% of the population of all candies. Thus, it is safe to assume that the trials are independent by the 10% condition.

Fixed number of trials: Since we selected 8 candies and the number of trials is also 8. Thus, the condition has been satisfied.

Probability of success: Since there are 20.5% chances for the candy being orange and the probability of success is also 20.5%. Thus, the condition has been satisfied.

(b)

To determine

Probability for getting 3 orange M&M’s.

Answer to Problem R6.5RE

Probability for getting 3 orange M&M’s is approx. 0.1532.

Explanation of Solution

Given information:

Number of trials, $n=8$

Probability of success, $p=20.5\%=0.205$

According to binomial probability definition,

  $P\left(X=k\right)=\left(\begin{array}{c}n\\ k\end{array}\right)\cdot p^k\cdot {\left(1-p\right)}^{n-k}$

At $k=3$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=3\right)=\left(\begin{array}{c}8\\ 3\end{array}\right)\cdot {\left(0.205\right)}^3\cdot {\left(1-0.205\right)}^{8-3}\\ =\frac{8!}{3!\left(8-3\right)!}\cdot {\left(0.205\right)}^3\cdot {\left(0.795\right)}^5\\ =56\cdot {\left(0.205\right)}^3\cdot {\left(0.795\right)}^5\\ \approx 0.1532\end{array}$

Thus,

Probability for getting 3 orange M&M’s is approx. 0.1532.

(c)

To determine

Probability for the event $X\ge 4$ .

Answer to Problem R6.5RE

Probability,

  $P\left(X\ge 4\right)=0.0610$

Explanation of Solution

Given information:

Number of trials, $n=8$

Probability of success, $p=20.5\%=0.205$

According to the binomial probability,

  $P\left(X=k\right)=\left(\begin{array}{c}n\\ k\end{array}\right)\cdot p^k\cdot {\left(1-p\right)}^{n-k}$

Addition rule for mutually exclusive event:

  $P\left(A\cup B\right)=P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)$

At $k=4$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=4\right)=\left(\begin{array}{c}8\\ 4\end{array}\right)\cdot {\left(0.205\right)}^4\cdot {\left(1-0.205\right)}^{8-4}\\ =\frac{8!}{4!\left(8-4\right)!}\cdot {\left(0.205\right)}^4\cdot {\left(0.795\right)}^4\\ \approx 0.0494\end{array}$

At $k=5$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=5\right)=\left(\begin{array}{c}8\\ 5\end{array}\right)\cdot {\left(0.205\right)}^5\cdot {\left(1-0.205\right)}^{8-5}\\ =\frac{8!}{5!\left(8-5\right)!}\cdot {\left(0.205\right)}^5\cdot {\left(0.795\right)}^3\\ \approx 0.0102\end{array}$

At $k=6$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=6\right)=\left(\begin{array}{c}8\\ 6\end{array}\right)\cdot {\left(0.205\right)}^6\cdot {\left(1-0.205\right)}^{8-6}\\ =\frac{8!}{6!\left(8-6\right)!}\cdot {\left(0.205\right)}^6\cdot {\left(0.795\right)}^2\\ \approx 0.0013\end{array}$

At $k=7$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=7\right)=\left(\begin{array}{c}8\\ 7\end{array}\right)\cdot {\left(0.205\right)}^7\cdot {\left(1-0.205\right)}^{8-7}\\ =\frac{8!}{7!\left(8-7\right)!}\cdot {\left(0.205\right)}^7\cdot {\left(0.795\right)}^1\\ \approx 0.0001\end{array}$

At $k=8$ ,

The binomial probability to be evaluated as:

  $\begin{array}{l}P\left(X=8\right)=\left(\begin{array}{c}8\\ 8\end{array}\right)\cdot {\left(0.205\right)}^8\cdot {\left(1-0.205\right)}^{8-8}\\ =\frac{8!}{8!\left(8-8\right)!}\cdot {\left(0.205\right)}^8\cdot {\left(0.795\right)}^0\\ \approx 0.0000\end{array}$

Since two different numbers of successes are impossible on same simulation.

Apply addition rule for mutually exclusive events:

  $\begin{array}{l}P\left(X\ge 4\right)=P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)+P\left(X=7\right)+P\left(X=8\right)\\ =0.0494+0.0102+0.0013+0.0001+0.0000\\ =0.0610\\ =6.10\%\end{array}$

Thus,

We get at least 4 orange candies among randomly selected 8 candies around 6.10% of the time.

(d)

To determine

Whether the result provides convincing evidence that Mars’s claim about its M&M’s is false.

Answer to Problem R6.5RE

There is no convincing evidence that Mars’s claim about its M&M’s is false.

Explanation of Solution

Given information:

Number of trials, $n=8$

Probability of success, $p=20.5\%=0.205$

From Part (c) result,

We have

  $P\left(X\ge 4\right)=0.0610$

When the probability is less than 0.05, it is considered to be small.

But note that

In this case, the probability is large, because 0.0610 is greater than 0.05.

This indicates

It is likely to obtain 4 or more orange candies.

Thus,

There is no convincing evidence that Mars’s claim about its M&M’s is false.