Two cards are drawn from a shuffled deck. What is the probability that both are red? If at least one is red, what is the probability that both are red? If at least one is a red ace, what is the probability that both are red? If exactly one is a red ace, what is the probability that both are red?

Two cards are drawn from a shuffled deck.

Probability is a metric for determining the possibility of an event occurring.

The probability is given as $P\left(R\right)=\frac{N\left(R\right)}{N\left(S\right)}$.

Two cards are drawn between 26 red cards so the number of elements of two red cards is given below.

$\begin{array}{c}N\left(R\right)=C(26,2)\\ N\left(R\right)=325\end{array}$

The number of elements in a sample space is given as follows.

$\begin{array}{c}N\left(S\right)=C(52,2)\\ N\left(S\right)=1326\end{array}$

Hence the probabilities are given below.

$\begin{array}{c}P\left(R\right)=\frac{326}{1326}\\ P\left(R\right)=\frac{25}{102}\end{array}$

The probability B of getting one red card is given as follows.

$\begin{array}{c}P\left(B\right)=1-\frac{C(26,2)}{C(52,2)}\\ P\left(B\right)=\frac{77}{102}\end{array}$

The probability of getting both red if at least one of them is red is given below.

$\begin{array}{c}{P}_B\left(R\right)=\frac{P\left(RB\right)}{P\left(B\right)}\\ P_B\left(R\right)=\frac{25/102}{77/102}\\ P_B\left(R\right)=\frac{25}{77}\end{array}$

The probability of getting red if E at least one is red is given below.

$P_E\left(R\right)=\frac{N\left(RE\right)}{N\left(E\right)}$

Let’s assume that the first card is an ace so the number of red ace cards are two and there are still 50 cards so each red will make a couple with the other 50 cards so it will be 100 couples and one other couple 2 red aces.

$\begin{array}{c}N\left(E\right)=2\times 50+1\\ N\left(E\right)=101\end{array}$

Among 101 couples, the couple where the first and second cards are red is 101 black cards with the first ace and another 26 cards with the second ace .

$\begin{array}{c}N\left(RE\right)=101-26-26\\ N\left(RE\right)=49\end{array}$

The value of probability becomes as follows.

$P_B\left(R\right)=\frac{49}{101}$

Find the probability for D= exactly there are only ace.

$\begin{array}{c}N\left(D\right)=2\times 50\\ N\left(D\right)=100\\ P\left(RD\right)=100-26-26\\ P\left(RD\right)=48\end{array}$

The probability is given below.

$\begin{array}{c}{P}_D\left(R\right)=\frac{P\left(DR\right)}{P\left(D\right)}\\ P_D\left(R\right)=\frac{48}{100}\\ P_D\left(R\right)=\frac{12}{25}\end{array}$

Hence the probability is mentioned below.

$\begin{array}{c}P\left(R\right)=\frac{25}{102}\\ P\left(B\right)=\frac{77}{102}\\ P_E\left(R\right)=\frac{25}{77}\\ P_B\left(R\right)=\frac{49}{101}\\ P_D\left(R\right)=\frac{12}{25}\end{array}$

$\begin{array}{l}p\left(A\right)=\left(\frac{n}{N}\right)\\ p\left(R\right)=\frac{3}{N}\end{array}$