Chapter 10.4, Problem 17E

Solution

(a)

To calculate: The probability of getting exactly one 6.

The probability of getting exactly one 6 is 0.39.

Given information:

The numbers of trails n is 4 and the numbers outcomes k is 1 and p is the probability of success.

Formula used:

The formula to calculate the binomial probability is given as,$P\left(X=K\right)=\left(\begin{array}{l}n\\ k\end{array}\right).p^k.{\left(1-p\right)}^{n-k}$   ...... (1)

Here, n is the numbers of trails, k is the numbers of outcomes and p is the probability of success.

Calculation:

Since, a dice is rolled, so there may be 6 possible outcomes that are 1,2,3,4,5 and 6.

So, the probability of success p is $\frac{1}{6}$

The number of successes among a fixed number of independent trials at a constant

Substitute 1 for k , $\frac{1}{6}$ for p and 4 for n in equation (1),

  $\begin{array}{c}P\left(X=1\right)=\left(\begin{array}{l}4\\ 1\end{array}\right)\cdot {\left(\frac{1}{6}\right)}^1\cdot {\left(1-\frac{1}{6}\right)}^{4-1}\\ =\frac{4!}{1!\left(4-1\right)!}\cdot {\left(\frac{1}{6}\right)}^1\cdot {\left(\frac{5}{6}\right)}^3\\ =6\cdot \frac{1}{6}\cdot {\left(\frac{5}{6}\right)}^3\\ =0.3858\end{array}$

Hence, the probability of getting exactly one 6 is 0.39.

(b)

To calculate: The probability of getting exactly two 6.

The probability of getting exactly two 6 is 0.12.

Given information:

The numbers of trails n is 4 and the numbers outcomes k is 2 and p is the probability of success.

Formula used:

The formula to calculate the binomial probability is given as,$P\left(X=K\right)=\left(\begin{array}{l}n\\ k\end{array}\right).p^k.{\left(1-p\right)}^{n-k}$ …… (1)

Here, n is the numbers of trails, k is the numbers of outcomes and p is the probability of success.

Calculation:

Since, a dice is rolled, so there may be 6 possible outcomes that are 1,2,3,4,5 and 6.

So, the probability of success p is $\frac{1}{6}$

The number of successes among a fixed number of independent trials at a constant

Substitute 2 for k , $\frac{1}{6}$ for p and 4 for n in equation (1),

  $\begin{array}{c}P\left(X=2\right)=\left(\begin{array}{l}4\\ 2\end{array}\right)\cdot {\left(\frac{1}{6}\right)}^2\cdot {\left(1-\frac{1}{6}\right)}^{4-2}\\ =\frac{4!}{2!\left(4-2\right)!}\cdot {\left(\frac{1}{6}\right)}^2\cdot {\left(\frac{5}{6}\right)}^2\\ =6\cdot {\left(\frac{1}{6}\right)}^2\cdot {\left(\frac{5}{6}\right)}^2\\ =\frac{150}{1296}\end{array}$

Solve further,

  $\begin{array}{c}P\left(X=2\right)=\frac{125}{324}\\ =0.1157\end{array}$

Hence, the probability of getting exactly two 6 is 0.12.

(c)

To calculate: The probability of getting at least two 6’s.

The probability of getting at least two 6’s is 0.13.

Given information:

The numbers of trails n is 4 and the numbers outcomes k is 0,1 and p is the probability of success.

Formula used:

The formula to calculate the binomial probability is given as,$P\left(X=K\right)=\left(\begin{array}{l}n\\ k\end{array}\right).p^k.{\left(1-p\right)}^{n-k}$   ...... (1)

Here, n is the numbers of trails, k is the numbers of outcomes and p is the probability of success.

Calculation:

Since, a dice is rolled, so there may be 6 possible outcomes that are 1,2,3,4,5 and 6.

So, the probability of success p is $\frac{1}{6}$

The number of successes among a fixed number of independent trials at a constant

Substitute 0 for k , $\frac{1}{6}$ for p and 4 for n in equation (1),

  $\begin{array}{c}P\left(X=0\right)=\left(\begin{array}{l}4\\ 0\end{array}\right)\cdot {\left(\frac{1}{6}\right)}^0\cdot {\left(1-\frac{1}{6}\right)}^{4-0}\\ =\frac{4!}{0!\left(4-0\right)!}\cdot {\left(\frac{1}{6}\right)}^0\cdot {\left(\frac{5}{6}\right)}^4\\ =1\cdot 1\cdot {\left(\frac{5}{6}\right)}^3\\ \end{array}$

Solve further,

  $\begin{array}{c}P\left(X=0\right)=\frac{625}{1296}\\ =0.4823\end{array}$

Substitute 1 for k , $\frac{1}{6}$ for p and 4 for n in equation (1),

  $\begin{array}{c}P\left(X=1\right)=\left(\begin{array}{l}4\\ 1\end{array}\right)\cdot {\left(\frac{1}{6}\right)}^1\cdot {\left(1-\frac{1}{6}\right)}^{4-1}\\ =\frac{4!}{1!\left(4-1\right)!}\cdot {\left(\frac{1}{6}\right)}^1\cdot {\left(\frac{5}{6}\right)}^3\\ =6\cdot \frac{1}{6}\cdot {\left(\frac{5}{6}\right)}^3\\ =0.3858\end{array}$

Now, Using the Addition rule for mutually exclusive events:

  $P(AorB)=P(A)+P(B)$

  $\begin{array}{c}P\left(X\le 1\right)=P(X=0)+P\left(X=1\right)\\ =0.4823+0.3858\\ =0.8681\end{array}$

Using the complement rule

  $P\left(\text{not}A\right)=1-P\left(X\le 1\right)$

  $\begin{array}{c}P\left(X\ge 2\right)=1-P(X\le 1)\\ =1-0.8681\\ =0.1319\end{array}$

Hence, the probability of getting at least two 6’s is 0.13.