Chapter 0.5, Problem 3E

(a)

To determine

To find: the probability of getting a club or diamond.

Answer to Problem 3E

The probability of club or diamond is $\frac{1}{2}$ .

Explanation of Solution

Given:

Total number of posssible outcomes of cards is $52$ .

Concept used:

The number of ways of arranging $n$ different item in order is $n_1\times n_2\times n_3\times \mathrm{....}{n}_n$ .

The probability formula is used to compute the probability of an event to occur as:

  $P\left(A\right)=\frac{\text{Number of favorable outcome}}{\text{Total Number of favorable outcomes}}$ .

Calculation:

Consider the standard deck of cards . the total number of posssible outcomes of cards is $52$ .

  $P\left(\text{club or diamond}\right)$ .

This situation is mutually exclusive because a card cannot be both a club and a diamond.

The probability of club or diamond is as follows:

  $P\left(\text{club or diamond}\right)=\frac{P\left(\text{club}\right)}{t\text{otal outcomes}}+\frac{P\left(\text{diamond}\right)}{\text{total outcomes}}$ .

Here,

Total outcomes of cards $=52$ .

  $\begin{array}{l}\frac{P\left(\text{club}\right)}{t\text{otal outcomes}}=\frac{13}{52}.\\ \frac{P\left(\text{diamond}\right)}{t\text{otal outcomes}}=\frac{13}{52}.\end{array}$

Substitute the values as follows:

  $\begin{array}{l}P\left(\text{club or diamond}\right)=\frac{13}{52}+\frac{13}{52}.\\ \Rightarrow \frac{26}{52}=\frac{1}{2}.\end{array}$

Hence, the probability of club or diamond is $\frac{1}{2}$ .

(b)

To determine

To find: the probability of getting ace or spade.

Answer to Problem 3E

The probability of ace or spade is $\frac{4}{13}$ .

Explanation of Solution

Given:

Total number of posssible outcomes of cards is $52$ .

Concept used:

The number of ways of arranging $n$ different item in order is $n_1\times n_2\times n_3\times \mathrm{....}{n}_n$ .

The probability formula is used to compute the probability of an event to occur as:

  $P\left(A\right)=\frac{\text{Number of favorable outcome}}{\text{Total Number of favorable outcomes}}$ .

Calculation:

Consider the following probability:

  $P\left(\text{ace or spade}\right)$

This situation is not mutually exclusive because there is an ace of spade. The probability of ace or spade is as follows:

  $P\left(\text{ace or spade}\right)=\frac{P\left(\text{ace}\right)}{t\text{otal outcomes}}+\frac{P\left(\text{spade}\right)}{\text{total outcomes}}-\frac{P\left(\text{ace and spade}\right)}{\text{total outcomes}}$ .

Here,

Total outcomes of dice $=6\times 6=36.$

  $\begin{array}{l}\frac{P\left(6\right)}{t\text{otal outcomes}}=\frac{5}{36}.\\ \frac{P\left(7\right)}{t\text{otal outcomes}}=\frac{6}{36}.\end{array}$

Substitute the values as follows:

  $\begin{array}{l}P\left(\text{ace or spade}\right)=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}.\\ \Rightarrow \frac{16}{52}=\frac{4}{13}.\end{array}$

Hence, the probability of ace or spade is $\frac{4}{13}$ .

(c)

To determine

To find: the probability of getting jack or red cards.

Answer to Problem 3E

The probability of jack or red card is $\frac{7}{13}$ .

Explanation of Solution

Given:

Total number of posssible outcomes of cards is $52$ .

Concept used:

The number of ways of arranging $n$ different item in order is $n_1\times n_2\times n_3\times \mathrm{....}{n}_n$ .

The probability formula is used to compute the probability of an event to occur as:

  $P\left(A\right)=\frac{\text{Number of favorable outcome}}{\text{Total Number of favorable outcomes}}$ .

Calculation:

Consider the following probability:

  $P\left(\text{jack or redcard}\right)$ .

This situation is not mutually exclusive because two jacks are red. The probability of jack or red card is as follows:

  $P\left(\text{jack or redcard}\right)=\frac{P\left(\text{jack}\right)}{\text{total outcomes}}+\frac{P\left(\text{red}\right)}{\text{total outcomes}}-\frac{P\left(\text{jack and red}\right)}{\text{total outcomes}}$ .

Here:

Total outcome of cards $=52$ .

  $\begin{array}{l}\frac{P\left(\text{jack}\right)}{\text{total outcomes}}=\frac{4}{52}.\\ \frac{P\left(\text{red}\right)}{\text{total outcomes}}=\frac{26}{52}.\\ \frac{P\left(\text{jack and red}\right)}{\text{total outcomes}}=\frac{2}{52}.\end{array}$

Substituting the values as follows:

  $P\left(\text{jack or redcard}\right)=\frac{P\left(\text{jack}\right)}{\text{total outcomes}}+\frac{P\left(\text{red}\right)}{\text{total outcomes}}-\frac{P\left(\text{jack and red}\right)}{\text{total outcomes}}$ .

Simplify it as follows:

  $\begin{array}{l}P\left(\text{jack or redcard}\right)=\frac{4}{52}+\frac{26}{52}-\frac{2}{52}.\\ \Rightarrow \frac{28}{52}=\frac{7}{13}.\end{array}$

Hence, the probability of jack or red card is $\frac{7}{13}$ .