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 true?
(a) $9P\left( A \right)=4P\left( H \right)$
(b) $P\left( S \right)=4P\left( A\cap H \right)$
(c) $3P\left( H \right)=3P\left( A\cup S \right)$
(d) None of these

Answer 1

We start solving the problem by recalling the suites, face cards, and remaining cards present in the pack of 52 cards. We then remove the required four cards and make note of the remaining cards present in the pack. We then find the probabilities of given events using the cards present in the remaining pack. We then check the given options with the obtained values to get the required result.

Complete step-by-step solution
According to the problem, we are given that the face cards and tens are removed from the pack of 52 cards, and a card is drawn at random from the remaining. If the events are defined as follows:
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, we need to find which of the given options are true.
We know that a pack of 52 cards contains four suits namely clubs, diamonds, hearts, and spades each containing 13 cards.
We know that the 13 cards contain Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, and king. We know that Jack, queen, and king are known as face cards.
According to the problem, face cards and 10 are removed from each suit of cards. So, we need to remove four cards from each suit which gives us a pack of 36 cards with 9 cards in each suit.
Now, let us find the probabilities of events A, H, and S.
So, we have $P\left( A \right)=\dfrac{\text{no}\text{. of ways of drawing ace from remaining 36 cards}}{\text{no}\text{. of ways of drawing a card from remaining 36 cards}}$.
$\Rightarrow P\left( A \right)=\dfrac{\text{4}}{\text{36}}$.
$\Rightarrow P\left( A \right)=\dfrac{\text{1}}{\text{9}}$ ---(1).
Now, we have $P\left( H \right)=\dfrac{\text{no}\text{. of ways of drawing a card belonging to hearts from remaining 36 cards}}{\text{no}\text{. of ways of drawing a card from remaining 36 cards}}$.
$\Rightarrow P\left( H \right)=\dfrac{9}{\text{36}}$.
$\Rightarrow P\left( H \right)=\dfrac{1}{\text{4}}$ ---(2).
Now, we have $P\left( S \right)=\dfrac{\text{no}\text{. of ways of drawing a card belonging to spades from remaining 36 cards}}{\text{no}\text{. of ways of drawing a card from remaining 36 cards}}$.
$\Rightarrow P\left( S \right)=\dfrac{9}{\text{36}}$.
$\Rightarrow P\left( S \right)=\dfrac{1}{\text{4}}$ ---(3).
Now, $A\cap H$ represents the event of drawing an ace which belongs to the hearts suite. Now, let us find the probability of the event $A\cap H$.
So, we have $P\left( A\cap H \right)=\dfrac{\text{no}\text{. of ways of drawing ace of hearts from remaining 36 cards}}{\text{no}\text{. of ways of drawing a card from remaining 36 cards}}$.
$\Rightarrow P\left( A\cap H \right)=\dfrac{1}{36}$ ---(4).
Now, $A\cup S$represents the event of drawing either an ace or card which belongs to the spade suit or both. Now, let us find the probability of the event $A\cup S$.
So, we have $P\left( A\cup S \right)=\dfrac{\text{no}\text{. of ways of drawing wither an ace or card from a spade or both from the remaining 36 cards}}{\text{no}\text{. of ways of drawing a card from remaining 36 cards}}$.
$\Rightarrow P\left( A\cap H \right)=\dfrac{12}{36}$.
$\Rightarrow P\left( A\cap H \right)=\dfrac{1}{3}$ ---(5).
From equations (1), let us consider $9P\left( A \right)$.
So, we have $9P\left( A \right)=1$.
$\Rightarrow 9P\left( A \right)=4\times \dfrac{1}{4}$.
From equation (2), we get $9P\left( A \right)=4P\left( H \right)$ ---(6).
We have $P\left( S \right)=\dfrac{1}{\text{4}}$.
$\Rightarrow P\left( S \right)=9\times \dfrac{1}{\text{36}}$.
$\Rightarrow P\left( S \right)=9P\left( A\cap H \right)$ ---(7).
Now, let us consider $3P\left( H \right)$.
$\Rightarrow 3P\left( H \right)=3\times \dfrac{1}{4}$.
$\Rightarrow 3P\left( H \right)=\dfrac{3}{4}$.
$\Rightarrow 3P\left( H \right)=\dfrac{9}{4}\times \dfrac{1}{3}$.
$\Rightarrow 3P\left( H \right)=\dfrac{9}{4}P\left( A\cup S \right)$ ---(8).
From equation (6), (7), and (8), we get that option (a) is true.
The correct option for the given problem is (a).

Note: We should make sure about the notations of representing the event before solving this problem. We should not make calculation mistakes while solving this problem. We should know that each contains an ace card before finding the probabilities of events $A\cap H$ and $A\cup S$. We can also use $P\left( A\cup S \right)=P\left( A \right)+P\left( S \right)-P\left( A\cap S \right)$ to find the probability of event $A\cup S$.