Chapter 13.5, Problem 24E

To determine

To calculate: The value of probability $P\left(\text{6 of clubs}\right)$ provided that the card is black.

Answer to Problem 24E

The value of probability $P\left(\text{6 of clubs}\right)$ provided that the card is black is $\frac{1}{26}$ .

Explanation of Solution

Given information:

A deck of 52 cards and the first card drawn is a black card.

Formula used:

Probability of an event Eis $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Given an event Q , the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

Calculation:

Consider the provided information that deck of 52 cards is there and the first card drawn is a black card.

The probability of $P\left(\text{6 of clubs}\right)$ provided that the card is black.

Let A be an event that card is black 6 of club and B be an event a black card is drawn.

There are 26 black cards and 1 black 6 of club card.

Recall that probability of an event Eis $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Probability of drawing a black card is,

  $P\left(B\right)=\frac{26}{52}$

Probability of drawing a black 6 of club is,

  $P\left(A\text{ and B}\right)=\frac{1}{52}$

Recall that given an event Q , the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

The probability of $P\left(\text{queen of hearts}\right)$ provided that the card is black.

  $\begin{array}{c}P\left(A|B\right)=\frac{\frac{1}{52}}{\frac{26}{52}}\\ =\frac{1}{26}\end{array}$

Thus, the value of probability $P\left(\text{6 of clubs}\right)$ provided that the card is black is $\frac{1}{26}$ .