Question

One card is drawn randomly from a well shuffled pack of \[52\] cards. Find the probability that the card is not a diamond.

Answer 1

Hint:We subtract the number of diamond cards from total number of cards. We use the concept of combinations to find the number of ways to choose one card from a total number of cards and then using the method for probability we find the probability of not a diamond 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 step-by-step answer:
We know a deck contains \[52\] cards where each suit has \[13\] cards in it
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 \[13\] diamonds out of the total number of cards.
We subtract number of cards that are diamond from total number of cards
So number of cards we can choose from is \[52 - 13 = 39\]
Now we choose one card from \[39\] cards
Number of ways to choose one card out of \[39\] cards is given by formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
Substituting the value for \[n = 39,r = 1\]
\[^{39}{C_1} = \dfrac{{39!}}{{38!1!}}\]
Now since we know factorial opens up as \[n! = n(n - 1)!\]
So, \[^{39}{C_1} = \dfrac{{39 \times 38!}}{{38!}} = 39\]
Now we find the probability of choosing a non-diamond card from a deck of \[52\] cards.
Probability is given by ways of choosing one card from \[39\] cards divided by probability of choosing one card from \[52\] cards.
Probability \[ = \dfrac{{39}}{{52}}\]
Writing the numerator and denominator in factored form
 Probability \[ = \dfrac{{13 \times 3}}{{13 \times 4}}\]
Cancel out the same terms from numerator and denominator.
Probability \[ = \dfrac{3}{4}\]
Thus, the probability of choosing one card from a deck of \[52\] cards such that the card is not a diamond is \[\dfrac{3}{4}\]or \[0.75\].

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.