Question

From a deck of $52$ cards, two cards are missing. One card is drawn at random from the remaining cards and found to be a spade. Find the probability that
$\left(1\right)$ both the missing cards were spades.
$\left(2\right)$ one of the missing card was spade.
$\left(3\right)$ neither of the missing cards was spade.
$\left(4\right)$ both the missing cards were same suit.

Answer 1

It is given that drawn card is spade. So, its probability is $1$. Hence, Probability that both the missing cards were spades is equivalent to drawing three cards from a full deck, such that the first card is a spade and then find the probability that the second and third cards are also spades,

$\left(1\right)$Therefore, probability that both the missing cards were spades is,
$P\left(\text{both spades}\right)=\frac{12}{51}\cdot \frac{11}{50}=\frac{22}{425}$

$\therefore$ Probability that both the missing cards were spades is $\frac{22}{425}.$

Similarly,
$\left(2\right)$ Probability that one of the missing card was spade is,
$P\left(\text{one spade}\right)=\frac{12}{51}\cdot \frac{39}{50}+\frac{39}{51}\cdot \frac{12}{50}=\frac{156}{425}$

$\therefore$ Probability that one of the missing card was spade is $\frac{156}{425}.$

$\left(3\right)$ Probability that neither of the missing cards was spade is,
$P\left(\text{no spade}\right)=\frac{39}{51}\cdot \frac{38}{50}=\frac{247}{425}$

$\therefore$ Probability that neither of the missing cards was spade is $\frac{247}{425}.$

$\left(4\right)$ Probability that both the missing cards were same suit is given as,
$P\left(\text{same suit}\right)=P\left(\text{both spades}\right)+P\left(\text{both clubs}\right)+P\left(\text{both diamonds}\right)+P\left(\text{both hearts}\right)$
$=\frac{12}{51}\cdot \frac{11}{50}+\frac{13}{51}\cdot \frac{12}{50}+\frac{13}{51}\cdot \frac{12}{50}+\frac{13}{51}\cdot \frac{12}{50}$
$=\frac{12}{51}\cdot \frac{11}{50}+3(\frac{13}{51}\cdot \frac{12}{50})$
$=\frac{4}{17}$

$\therefore$ Probability that both the missing cards were same suit is $\frac{4}{17}.$