Question

A deck of cards contains 10 black cards and 10 red cards. The first card drawn is not replaced before drawing the second card.What is the probability of selecting a red card followed by a red card? 141019519938

Answer 1

Solution:

The probability of an event is expressed as

$P(event)=\frac{number\text{ of desirable outcome}}{number\text{ of possible outcome}}$

Given:

$\begin{gathered} number\text{ of black cards = 10} \\ number\text{ of red cards = 10} \\ Total\text{ number of cards = 20} \end{gathered}$

In this case, the number of possible outcomes is 20.

Given that the first card is drawn and not replaced before drawing the second card, the probability of selecting a red card followed by a red card is expressed as

$P(red\text{ and red\rparen=P\lparen first red\rparen}\times P(second\text{ red without replacement\rparen}$

For the first red, we have

$\begin{gathered} P(first\text{ red\rparen=}\frac{number\text{ of red cards}}{number\text{ of cards\lparen number of possible outcomes\rparen}}=\frac{10}{20} \\ \Rightarrow P(first\text{ red\rparen=}\frac{1}{2} \end{gathered}$

For the second card, we have

$\begin{gathered} P(second\text{ red\rparen=}\frac{number\text{ of red cards left}}{number\text{ of cards left}}=\frac{10-1}{20-1} \\ =\frac{9}{19} \end{gathered}$

Thus,

$\begin{gathered} P(\text{ first red and second red\rparen=}\frac{1}{2}\times\frac{9}{19} \\ \Rightarrow P(\text{ first red and second red\rparen=}\frac{9}{39} \end{gathered}$

Hence, the probability of selecting a red card followed by a red card is evaluated to be

$\frac{9}{38}$

The fourth option is the correct answer.