What is the probability of the second card being red after we know that the first card is not a heart ?? I have a hard time understanding this simple question!!
From a deck of 52 cards. Two cards are drawn randomly, one at a time without replacement. What is the probability of the second card being red after we know that the first card is not a heart ??
 A: $$\mathbb{P}[\text{Second card is Red}|\text{ First card is not Heart}]=$$
$$=\frac{\mathbb{P}[\text{Second card is Red}\cap\text{ First card is not Heart}]}{\mathbb{P}[\text{ First card is not Heart}]}=$$
$$=\frac{\frac{13}{52}\times \frac{25}{51}+\frac{26}{52}\times \frac{26}{51}}{\frac{39}{52}}\approx 0.5033$$
A: Let the probability you seek be denoted by $P$. Here is a geometric argument:

*

*Assume that the first card was red. Then you have $p_1 := \dfrac{25}{51}$ probability of picking a red card on the second go. One would expect $p_1 < P$.

*Assume the first card was black. Then the probability of picking a red card on the second choice is $p_2 := \dfrac{26}{51}$. The expectation is that $P < p_2$.

As the first card was not hearts (which is red), we would expect $P$ to be $\dfrac{2}{3}$ the way from $p_1$ to $p_2$. So:
\begin{align}
P &= p_1 + \dfrac{2}{3} (p_2 - p_1) =\\
  &= \dfrac{25}{51} + \dfrac{2}{3}\dfrac{1}{51} =\\
  &= \dfrac{77}{153} \approx 0.50327.
\end{align}
Edit (response to comment)
The $\dfrac{2}{3}$ is essentially a weight on assumption 2 while $\dfrac{1}{3}$ is the weight on assumption 1 given that the first card is not hearts. If hearts is ruled out, there are $3$ remaining possible classes of cards. Two of these are black; hence the weight $\dfrac{2}{3}$. Note that you could also write
$$P = \dfrac{1}{3}p_1 + \dfrac{2}{3}p_2$$
to, of course, arrive at the same end result.
A: I am denoting the events in which 1st card is diamond or (spade or club) as A,B respectively. 
P(A)=$\frac{1}{4}$ , P(B)=$\frac{1}{2}$ 
denoting the event in which 2nd card is red as C 
P$\left(\frac{C}{A}\right)$=$\frac{25}{51}$ 
P$\left(\frac{C}{B}\right)$=$\frac{26}{51}$ 
P(C∩(A∪B))=$\frac{1}{4}$×$\frac{25}{51}$+$\frac{1}{2}$
×$\frac{26}{51}$ 
P($\frac{C}{A∪B})$ =$\frac{\frac{1}{4}×\frac{25}{51}+\frac{1}{2}
×\frac{26}{51}}{{\frac{3}{4}}}$
A: $$P\left(R_{2}\cap H_{1}^{c}\right)=P\left(R_{2}\right)-P\left(R_{2}\cap H_{1}\right)=P\left(R_{2}\right)-P\left(R_{2}\mid H_{1}\right)P\left(H_{1}\right)=\frac{1}{2}-\frac{25}{51}\frac{1}{4}$$
So that:
$$P\left(R_{2}\mid H_{1}^{c}\right)=\frac{P\left(R_{2}\cap H_{1}^{c}\right)}{P\left(H_{1}^{c}\right)}=\frac{\frac{1}{2}-\frac{25}{51}\frac{1}{4}}{\frac{3}{4}}=\frac{77}{153}\approx0.503268$$
