Chapter 12.6, Problem 11CFU

(a)

To determine

To find:the number of possible sets of heads and tails that has $0$ heads.

Answer to Problem 11CFU

There is only one set that will have $0$ heads.

Explanation of Solution

Given: A coin is flipped five times.

Concept used:

Using binomial or pascal triangle the possible sets in the form of equation can be expanded.

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\frac{n\left(n-1\right)}{1.2}{x}^{n-2}{y}^2+\frac{n\left(n-1\right)\left(n-2\right)}{\mathrm{1.2.4}}{x}^{n-3}{y}^3+\mathrm{....}{y}^n$ .

Where $x=\text{head, }y=\text{tail}\text{.}$

Coin is tossed or flipped $5$ times so, the maximum value of $n=5$ .

Calculation:

Let $h$ represent the heads and $t$ represents tails. To find the number of possible sets, expand ${\left(h+t\right)}^5$ .

Use the sixth row of pascals triangle for the expansion.

  $h^5+5h^{4}t+10h^3{t}^2+10h^2{t}^3+5h^4+t^5$ .

To have $0$ heads, the only one possibility is all are tails.

The coefficient of $t^5$ .

Hence, there is only one set that will have $0$ heads.

(b)

To determine

To find:the number of possible sets of heads and tails that has $2$ heads.

Answer to Problem 11CFU

The possible sets are $10$ .

Explanation of Solution

Given:

A coin is flipped five times.

Concept used:

Using binomial or pascal triangle the possible sets in the form of equation can be expanded.

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\frac{n\left(n-1\right)}{1.2}{x}^{n-2}{y}^2+\frac{n\left(n-1\right)\left(n-2\right)}{\mathrm{1.2.4}}{x}^{n-3}{y}^3+\mathrm{....}{y}^n$ .

Where $x=\text{head, }y=\text{tail}\text{.}$

Coin is tossed or flipped $5$ times so, the maximum value of $n=5$ .

the number of head and tail is represented by the power of $h$ and $t$ respectively and the possible sets is the coefficient of the term that represents the number of head and number of tails.

Calculation:

Let $h$ represent the heads and $t$ represents tails. To find the number of possible sets, expand ${\left(h+t\right)}^5$ .

Use the sixth row of pascals triangle for the expansion.

  $h^5+5h^{4}t+10h^3{t}^2+10h^2{t}^3+5h^4+t^5$ .

To have the $2$ heads,the possible sets is equal to the coefficient of $h^2{t}^3$ .

since, the coefficient of $h^2{t}^3$ is $10$ .

Hence, the possible sets are $10$ .

(c)

To determine

To find:the number of possible sets of heads and tails that has at least $4$ heads.

Answer to Problem 11CFU

There is $6$ possible sets of head and tails in which there are at least $4$ heads.

Explanation of Solution

Given:

A coin is flipped five times.

Concept used:

Using binomial or pascal triangle the possible sets in the form of equation can be expanded.

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\frac{n\left(n-1\right)}{1.2}{x}^{n-2}{y}^2+\frac{n\left(n-1\right)\left(n-2\right)}{\mathrm{1.2.4}}{x}^{n-3}{y}^3+\mathrm{....}{y}^n$ .

Where $x=\text{head, }y=\text{tail}\text{.}$

Coin is tossed or flipped $5$ times so, the maximum value of $n=5$ .

the number of head and tail is represented by the power of $h$ and $t$ respectively and the possible sets is the coefficient of the term that represents the number of head and number of tails.

Calculation:

Let $h$ represent the heads and $t$ represents tails. To find the number of possible sets, expand ${\left(h+t\right)}^5$ .

Use the sixth row of pascals triangle for the expansion.

  $h^5+5h^{4}t+10h^3{t}^2+10h^2{t}^3+5h^4+t^5$ .

To have the at least $4$ heads means, there can be $4\text{ or 5}$ heads.

The total number of sets is the same as the sum of coefficient of $h^{4}t\text{ and }{h}^5$ .

Since, the sum of coefficient of $h^{4}t\text{ and }{h}^5$ is $1+5=6$ .

Hence, there is $6$ possible sets of head and tails in which there are at least $4$ heads.

(d)

To determine

To find:the number of possible sets of heads and tails that has at least $3$ heads.

Answer to Problem 11CFU

There is $26$ possible sets of head and tails in which there are at most $3$ heads.

Explanation of Solution

Given:

A coin is flipped five times.

Concept used:

Using binomial or pascal triangle the possible sets in the form of equation can be expanded.

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\frac{n\left(n-1\right)}{1.2}{x}^{n-2}{y}^2+\frac{n\left(n-1\right)\left(n-2\right)}{\mathrm{1.2.4}}{x}^{n-3}{y}^3+\mathrm{....}{y}^n$ .

Where $x=\text{head, }y=\text{tail}\text{.}$

Coin is tossed or flipped $5$ times so, the maximum value of $n=5$ .

the number of head and tail is represented by the power of $h$ and $t$ respectively and the possible sets is the coefficient of the term that represents the number of head and number of tails.

Calculation:

Let $h$ represent the heads and $t$ represents tails. To find the number of possible sets, expand ${\left(h+t\right)}^5$ .

Use the sixth row of pascals triangle for the expansion.

  $h^5+5h^{4}t+10h^3{t}^2+10h^2{t}^3+5h^4+t^5$ .

To have the at most $3$ heads means, there could be $\text{0,1,2 or 3}$ heads.

The total number of sets is the same as the sum of coefficient of $t^5,h{t}^4,h^2{t}^3\text{ and }{h}^3{t}^2.$

Since, the sum of coefficient of $t^5,h{t}^4,h^2{t}^3\text{ and }{h}^3{t}^2$ is $1+5+10+10=26$ .

Hence, there is $26$ possible sets of head and tails in which there are at most $3$ heads.