Question

From a pack of $52$ playing cards, face cards and tens are removed and kept aside then a card is drawn at random from the remaining cards. If
A : The event that the card drawn is an ace
H : The event that the card drawn is a heart
S : The event that the card drawn is a spade
then which of the following holds?

• A. $9P\left(A\right)=4P\left(H\right)$
• B. $P\left(S\right)=4P(A\cap H)$
• C. $3P\left(H\right)=3P(A\cup S)$
• D. $Noneofthese$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $9P\left(A\right)=4P\left(H\right)$
C $3P\left(H\right)=3P(A\cup S)$
D $Noneofthese$

Given from a pack of 52 cards, face cards and tens are removed.
$\Rightarrow$Total no. of remaining cards$=52-(12+4)=36$ cards
$A$: The event that the card drawn is an ace, $P\left(A\right)$=$\frac{4}{36}=\frac{1}{9}$

$H$: The event that the card drawn is an Heart, $P\left(H\right)$=$\frac{9}{36}=\frac{1}{4}$

$S$:The event that the card drawn is a Spade, $P\left(S\right)$=$\frac{9}{36}=\frac{1}{4}$
and $P(A\cap S)=\frac{1}{36}$(Since only 1 card exist which is spade and ace)
$P(A\cup S)=P\left(A\right)+P\left(S\right)-P(A\cap S)=\frac{1}{9}+\frac{1}{4}-\frac{1}{36}$ (since $P(A\cap S)=\frac{1}{36})$
$=$$\frac{1}{3}$
$\Rightarrow 9P\left(A\right)=4P\left(H\right)=3P(A\cup S)=1$ and 9$P(A\cap S)=P\left(H\right)$