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**Hint:**Probability of an event is the ratio of the number of outcomes favourable to the event, to the total number of outcomes.

Event is the occurrence whose probability has to be found.

In a well shuffled deck of 52 playing card there are:

Each groups of 13 cards is consist of: 3 Face cards: King, Queen, Jack

1 Ace & numbers from 2 to 10.

Therefore, there are 12 Face cards in total (6 red and 6 black), 4 Aces. 4 Kings (2 red and 2 black), 4 Queens (2 red and 2 black).

**Complete step by step solution:**Step 1

(i) Number of face card in well shuffled deck of 52 playing cards = 12

Step 2

$\because $ number of non-face card in well shuffled deck of 52 playing card = 52-12

= 40

Step 3

Probability ( a non-face card ) $ = \dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total number of outcomes}}}}$

$ = \dfrac{{40}}{{52}}$

$ = \dfrac{{10}}{{13}}$

Step 4

(ii) Number of black king in well shuffled deck of 52 playing cards = 2

Step 5

Number of red queen in well shuffled deck of 52 playing cards = 2

Step 6

Probability ( an event ) $ = \dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total number of outcomes}}}}$

Probability ( a black king ) $ = \dfrac{2}{{52}}$

Probability ( a red queen ) $ = \dfrac{2}{{52}}$

Step 7

Probability ( a black king or a red queen ) $ = \dfrac{2}{{52}} + \dfrac{2}{{52}}$

$ = \dfrac{4}{{52}}$

$ = \dfrac{1}{{13}}$

__Therefore, the probability of finding a non-face card in a well shuffled deck of 52 playing cards is $\dfrac{{10}}{{13}}$ . The probability of finding either a black king or a red queen in a well shuffled deck of 52 playing cards is $\dfrac{1}{{13}}$.__

**Note:**

The alternate probability calculations can also be done using combinations $C\left( {n,r} \right)$

(i) Probability ( a non-face card ) $ = \dfrac{{{\text{numbers of ways of selecting one non - face card among all non - face cards}}}}{{{\text{numbers of ways of selecting one card among a deck of well shuffled card }}}}$

\[ = \dfrac{{C\left( {40,1} \right)}}{{C\left( {52,1} \right)}}\]

=\[

= \dfrac{{\dfrac{{40!}}{{\left( {40 - 1} \right)!1!}}}}{{\dfrac{{52!}}{{\left( {52 - 1} \right)!1!}}}} \\

= \dfrac{{\dfrac{{40!}}{{39!}}}}{{\dfrac{{52!}}{{51!}}}} = \dfrac{{\dfrac{{40 \times {{39!}}}}{{{{39!}}}}}}{{\dfrac{{52 \times {{51!}}}}{{{{51!}}}}}} \\

= \dfrac{{40}}{{52}} \\

= \dfrac{{10}}{{13}} \\

\]

(ii) Probability (a black king or a red queen )$ = \dfrac{

{\text{numbers of ways of selecting one black king card among all black king cards }} \\

{\text{ + number of ways of selecting a red queen card among all red queen cards}} \\

}{{{\text{numbers of ways of selecting one cards among a deck of well shuffled card }}}}$

\[ = \dfrac{{C\left( {2,1} \right) + C\left( {2,1} \right)}}{{C\left( {52,1} \right)}}\]

\[ = \dfrac{{\dfrac{{2!}}{{\left( {2 - 1} \right)!1!}} + \dfrac{{2!}}{{\left( {2 - 1} \right)!1!}}}}{{\dfrac{{52!}}{{\left( {52 - 1} \right)!1!}}}}\]

\[ = \dfrac{{\dfrac{{2!}}{{1!1!}} + \dfrac{{2!}}{{1!1!}}}}{{\dfrac{{52!}}{{51!1!}}}}\]

\[

= \dfrac{{2 + 2}}{{\dfrac{{52 \times {{51!}}}}{{{{51!}}}}}} \\

= \dfrac{4}{{52}} \\

= \dfrac{1}{{13}} \\

\]

Probability of any event is always “greater than and equal to 0” and “less than and equal to 1”.

If there was to calculate Probability of both a red queen and a black king, then the probability of finding a black king card and a red queen card will get multiplied.

Students are likely to get confused with the fact that ace is included in face card or not, but keep in mind face cards are the cards that are completely covered with a picture, say of king, queen etc.

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