Chapter 10.4, Problem 18E

Solution

(a)

To calculate: The probability of getting exactly 5 heads.

The probability of getting exactly 5 head is 0.24.

Given information:

The numbers of trails n is 10 and the numbers outcomes k is 5 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 coin is tossed, so there may be two possible outcomes that are either head or tails.

So, the probability of success p is $\frac{1}{2}$

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

Substitute 5 for k , $\frac{1}{2}$ for p and 10 for n in equation (1),

  $\begin{array}{c}P\left(X=5\right)=\left(\begin{array}{l}10\\ 5\end{array}\right)\cdot {\left(\frac{1}{2}\right)}^5\cdot {\left(1-\frac{1}{2}\right)}^{10-5}\\ =\frac{10!}{5!\left(10-5\right)!}\cdot {\left(\frac{1}{2}\right)}^5\cdot {\left(\frac{1}{2}\right)}^5\\ =252\cdot {\left(\frac{1}{2}\right)}^{10}\\ \end{array}$

Solve further,

  $\begin{array}{c}P\left(X=5\right)=\frac{252}{1024}\\ =\frac{63}{256}\\ =0.2461\end{array}$

Hence, the number of getting exactly 5 head is 0.24.

(b)

To calculate: The probability of getting exactly 8 heads.

The probability of getting exactly 8 head is 0.044.

Given information:

The numbers of trails n is 10 and the numbers outcomes k is 8 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 coin is tossed, so there may be two possible outcomes that are either head or tails.

So, the probability of success p is $\frac{1}{2}$

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

Substitute 8 for k , $\frac{1}{2}$ for p and 10 for n in equation (1),

  $\begin{array}{c}P\left(X=8\right)=\left(\begin{array}{l}10\\ 8\end{array}\right)\cdot {\left(\frac{1}{2}\right)}^8\cdot {\left(1-\frac{1}{2}\right)}^{10-8}\\ =\frac{10!}{8!\left(10-8\right)!}\cdot {\left(\frac{1}{2}\right)}^8\cdot {\left(\frac{1}{2}\right)}^2\\ =45\cdot {\left(\frac{1}{2}\right)}^{10}\\ \end{array}$

Solve further,

  $\begin{array}{c}P\left(X=8\right)=\frac{45}{1024}\\ =0.04394\end{array}$

Hence, the number of getting exactly 8 head is 0.044.

(c)

To calculate: The probability of getting at least 8 heads.

The probability of getting at least 8 head is 0.055.

Given information:

The numbers of trails n is 10 and the numbers outcomes k is 8,9,10 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 coin is tossed, so there may be two possible outcomes that are either head or tails.

So, the probability of success p is $\frac{1}{2}$

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

Substitute 8 for k , $\frac{1}{2}$ for p and 10 for n in equation (1),

  $\begin{array}{c}P\left(X=8\right)=\left(\begin{array}{l}10\\ 8\end{array}\right)\cdot {\left(\frac{1}{2}\right)}^8\cdot {\left(1-\frac{1}{2}\right)}^{10-8}\\ =\frac{10!}{8!\left(10-8\right)!}\cdot {\left(\frac{1}{2}\right)}^8\cdot {\left(\frac{1}{2}\right)}^2\\ =45\cdot {\left(\frac{1}{2}\right)}^{10}\end{array}$

Solve further,

  $\begin{array}{c}P\left(X=8\right)=\frac{45}{1024}\\ =0.04394\end{array}$

Substitute 9 for k , $\frac{1}{2}$ for p and 10 for n in equation (1),

  $\begin{array}{c}P\left(X=9\right)=\left(\begin{array}{l}10\\ 9\end{array}\right)\cdot {\left(\frac{1}{2}\right)}^9\cdot {\left(1-\frac{1}{2}\right)}^{10-9}\\ =\frac{10!}{9!\left(10-9\right)!}\cdot {\left(\frac{1}{2}\right)}^9\cdot {\left(\frac{1}{2}\right)}^1\\ =10\cdot {\left(\frac{1}{2}\right)}^{10}\end{array}$

Solve further,

  $\begin{array}{c}P\left(X=9\right)=\frac{10}{1024}\\ =0.009765\end{array}$

Substitute 10 for k , $\frac{1}{2}$ for p and 10 for n in equation (1),

  $\begin{array}{c}P\left(X=10\right)=\left(\begin{array}{l}10\\ 10\end{array}\right)\cdot {\left(\frac{1}{2}\right)}^{10}\cdot {\left(1-\frac{1}{2}\right)}^{10-10}\\ =\frac{10!}{10!\left(10-10\right)!}\cdot {\left(\frac{1}{2}\right)}^9\cdot {\left(\frac{1}{2}\right)}^1\\ ={\left(\frac{1}{2}\right)}^{10}\end{array}$

Solve further,

  $\begin{array}{c}P\left(X=10\right)=\frac{1}{1024}\\ =0.0009765\end{array}$

Now, Using the Addition rule for mutually exclusive events:

  $P(AorB)=P(A)+P(B)$

  $\begin{array}{c}P\left(X\ge 8\right)=P(X=8)+P\left(X=9\right)+P\left(X=10\right)\\ =0.04394+0.009765+0.009765\\ =0.05468\end{array}$

Hence, the probability of getting at least 8 head is 0.055.