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Find the probability of drawing a king (one pick) from a shuffled standard deck of 52 cards. (A standard deck of cards is the most common type of deck used in most card games containing 52 cards)

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Last updated date: 20th Jun 2024
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Answer
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Hint: We use the concept of combinations to find the number of ways to choose one card from total number of cards and then using the method for probability we find the probability of a king card.
* Combination is given by \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
* Probability of an event is given by the number of possibilities divided by total number of possibilities.

Complete answer:
We know a deck contains \[52\]cards where there are 4 kings
Number of ways to choose one card out of \[52\]cards is given by formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
Substituting the value for \[n = 52,r = 1\]
\[^{52}{C_1} = \dfrac{{52!}}{{51!1!}}\]
Now since we know factorial opens up as \[n! = n(n - 1)!\]
So, \[^{52}{C_1} = \dfrac{{52 \times 51!}}{{51!}} = 52\]
Now there are 4 king cards out of the total number of cards.
Number of ways to choose one card out of 4 cards is given by formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
Substituting the value for \[n = 4,r = 1\]
\[^4{C_1} = \dfrac{{4!}}{{3!1!}}\]
Now since we know factorial opens up as \[n! = n(n - 1)!\]
So, \[^4{C_1} = \dfrac{{4 \times 3!}}{{3!}} = 4\]
Now we find the probability of choosing a king card from a deck of \[52\]cards.
Probability is given by dividing ways of choosing one card from 4 cards divided by probability of choosing one card from \[52\]cards.
Probability \[ = \dfrac{4}{{52}}\]
Writing the numerator and denominator in factored form
 Probability \[ = \dfrac{4}{{13 \times 4}}\]
Cancel out the same terms from numerator and denominator.
Probability \[ = \dfrac{1}{4}\]

Thus, probability of choosing one card from deck of \[52\] cards such that the card is a king is \[\dfrac{1}{4}\]or 0.25

Note:
Students should always check their answer of probability should be less than or equal to one and greater than or equal to zero. Students many times try to solve the combination formula by opening the factorial but that makes the solution complex instead try to cancel out as many factorial terms as you can from numerator and denominator.