Chapter 17, Problem 18E

(a)

To determine

To find: The probability of the 1st bull's eye researcher coming to the 3rd arrow.

Answer to Problem 18E

0.032

Explanation of Solution

Given:

An Olympic archer is 80 percent of the time able to hit the bull 's eye. The number of shots she fires is 6. Each shot is independent. The trials, therefore, are the Bernoulli trials.

Calculation:

The probability of the 1st bull's eye researcher coming to the 3rd arrow is,

  $\begin{array}{c}P\left(X=3\right)={\left(0.2\right)}^2\left(0.8\right)\\ =0.032\end{array}$

Thus, there is 3.2% possibility that the 1st bull's eye researcher coming to the 3rd arrow.

(b)

To determine

To find: On the 4th or 5th arrow, the probability that the researcher misses the bull 's eye.

Answer to Problem 18E

0.7379

Explanation of Solution

Given:

An Olympic archer is 80 percent of the time able to hit the bull 's eye. The number of shots she fires is 6. Each shot is independent. The trials, therefore, are the Bernoulli trials.

Calculation:

The probability that the researcher misses the bull 's eye is,

  $\begin{array}{c}P\left(\text{misses atleast once}\right)=1-P\left(\text{never missed}\right)\\ =1-\left({}^6{C}_0\right){\left(0.2\right)}^0{\left(0.8\right)}^6\\ =1-\frac{6!}{6!0!}{\left(0.8\right)}^6\\ =1-{\left(0.8\right)}^6\\ =1-0.2621\\ =0.7379\end{array}$

Therefore, there is 73.79% possibility that the researcher misses the bull 's eye.

(c)

To determine

To find: The probability that the first bull 's eye of a researcher would appear on the 4th or 5th arrow.

Answer to Problem 18E

0.0077

Explanation of Solution

Given:

An Olympic archer is 80 percent of the time able to hit the bull 's eye. The number of shots she fires is 6. Each shot is independent. The trials, therefore, are the Bernoulli trials.

Calculation:

 The probability that the first bull 's eye of a researcher would appear on the 4th or 5th arrow.

  $\begin{array}{c}P\left(X=4\right)+P\left(X=5\right)={\left(0.2\right)}^3\left(0.8\right)+{\left(0.2\right)}^4\left(0.8\right)\\ =0.0064+0.0013\\ =0.0077\end{array}$

Thus, there is 0.77% possibility that the first bull 's eye of a researcher would appear on the 4th or 5th arrow.

(d)

To determine

To find: the probability that researcher obtained exactly four bull’s eyes.

Answer to Problem 18E

0.2458

Explanation of Solution

Given:

An Olympic archer is 80 percent of the time able to hit the bull 's eye. The number of shots she fires is 6. Each shot is independent. The trials, therefore, are the Bernoulli trials.

Calculation:

The Probability that researcher obtained exactly four bull’s eyes is,

  $\begin{array}{c}P\left(X=4\right)=\left({}^6{C}_4\right)\times {\left(0.8\right)}^4{\left(0.2\right)}^2\\ =15\times 0.4096\times 0.04\\ =0.2458\end{array}$

Therefore, there is 24.58% possibility that researcher obtained exactly four bull’s eyes.

(e)

To determine

To find: the probability that researcher obtained minimum four bull’s eyes

Answer to Problem 18E

0.9011

Explanation of Solution

Given:

An Olympic archer is 80 percent of the time able to hit the bull 's eye. The number of shots she fires is 6. Each shot is independent. The trials, therefore, are the Bernoulli trials.

Calculation:

The probability that researcher obtained minimum four bull’s eyes

  $\begin{array}{c}P\left(X\ge 4\right)=P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)\\ =\left({}^6{C}_4\right)\times {\left(0.8\right)}^4{\left(0.2\right)}^2+\left({}^6{C}_5\right)\times {\left(0.8\right)}^5{\left(0.2\right)}^1+\left({}^6{C}_6\right)\times {\left(0.8\right)}^6{\left(0.2\right)}^0\\ =0.2458+0.3932+0.2621\\ =0.9011\end{array}$

Therefore, there is 90.11% chance that she gets at least 4 bull’s eyes.

(f)

To determine

To find: the probability that researcher obtained at most four bull’s eyes

Answer to Problem 18E

0.3447

Explanation of Solution

Given:

An Olympic archer is 80 percent of the time able to hit the bull 's eye. The number of shots she fires is 6. Each shot is independent. The trials, therefore, are the Bernoulli trials.

Calculation:

The Probability that she gets researcher obtained at most four bull’s eyes is

  $\begin{array}{c}P\left(X\ge 4\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)\\ =\left({}^6{C}_0\right)\times {\left(0.8\right)}^0{\left(0.2\right)}^{6-0}+\left({}^6{C}_1\right)\times {\left(0.8\right)}^1{\left(0.2\right)}^{6-1}+\left({}^6{C}_2\right)\times {\left(0.8\right)}^2{\left(0.2\right)}^{6-2}\\ +\left({}^6{C}_3\right)\times {\left(0.8\right)}^3{\left(0.2\right)}^{6-3}+\left({}^6{C}_4\right)\times {\left(0.8\right)}^4{\left(0.2\right)}^{6-4}\\ =0.0001+0.0015+0.0154+0.0819+0.2458\\ =0.3447\end{array}$

Thus, there 34.47% possibility that researcher obtained at most four bull’s eyes.