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**Hint:**In this question, we will use the formula for finding the probability which is given as:

$P(E) = \dfrac{{{\text{Total number of favorable events}}}}{{{\text{Total number of possible events}}}}$. First we will calculate the total number of possible events. For each case, we will find the number of favorable events and then use the formula.

**Complete step-by-step answer:**

In a deck of 52 cards, we are removing 2 red queens (one of hearts and other of diamond) and 2 black jacks (one of spade and other of club) are also being removed. Total card removed = 4.

So, cards remaining in deck = 48

Or we can say n(s) = 48 = Total number of possible events.

Now we need to find probability that 1 card is drawn from the remaining cards after reshuffling is:

(a) a king

There are 4 kings in a pack of 52 cards.

$\therefore $n(${E_1}$) = 4 = total number of favorable events.

Let ${E_1}$be the event that the card drawn is a king.

$P({E_1}) = \dfrac{{{\text{Total number of favorable events}}}}{{{\text{Total number of possible events}}}} = \dfrac{{n({E_1})}}{{n(s)}} = \dfrac{4}{{48}} = \dfrac{1}{{12}}$.

(b) of red color

There are in total 26 red cards, 13 of hearts and 13 of diamonds but we have removed 2 red queens therefore, the total remaining red cards is 24.

Now let${E_2}$ be the probability that the drawn card is red.

$\therefore $ n(${E_2}$)=24

$P({E_2}) = \dfrac{{n({E_2})}}{{n(s)}} = \dfrac{{24}}{{48}} = \dfrac{1}{2}$.

(c) a face card

There are in total 12 face cards. Now it is being told that we have removed 2 queens and 2 jacks which are also face cards hence the remaining face cards will be 8.

Now, let${E_3}$be the event of obtaining a face card while making a draw.

$\therefore $ n(${E_3}$)=8

$P({E_3}) = \dfrac{{n({E_3})}}{{n(s)}} = \dfrac{8}{{48}} = \dfrac{1}{6}$.

(c) a queen

Number of queens in a deck of 52 cards = 4 but 2 queens have been removed.

$\therefore $ n(${E_4}$)=2, where ‘${E_4}$’ is the event of occurring a queen.

$P({E_4}) = \dfrac{{n({E_4})}}{{n(s)}} = \dfrac{2}{{48}} = \dfrac{1}{{24}}$.

**Note:**Solving such probability questions simply requires understanding of what is being removed and how many cards are being left behind. Always remember there are only 12 face cards and ace is not a face card. We need to remember that there are in total 4 houses that are heart, diamond, and spade and club each having 13 cards in them.

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