Question

A card is drawn from a pack of cards. Find the probability that the card will be a queen or a heart
A. \[\dfrac{4}{3}\]
B. \[\dfrac{{16}}{3}\]
C. \[\dfrac{4}{{13}}\]
D. \[\dfrac{5}{3}\]

Answer 1

Hint: In our case, we have to determine the probability of receiving either a queen or a heart. The total number of cards in the pack will be determined first. The number of queens and hearts in the pack will then be determined. Finally, in order to calculate the necessary probability, we shall apply the probability formula.



Formula Used:Event’s probability formula can be used:
\[P(E) = \dfrac{{{\rm{Number of favourable outcomes}}}}{{{\rm{Number of total outcomes }}}}\]



Complete step by step solution:We are provided in the data that a card is drawn from the pack of cards.
And asked to determine the probability that the card will be a queen or a heart
We have been already known that total number of card in deck is
\[52\] Cards
So, one card can be drawn from the pack of 52 cards in
\[^{52}{C_1} = 52\] Ways
Therefore, we have
\[{\rm{n}}({\rm{S}}) = 52\]
Also we know that the total number of Heart Cards in a deck = \[13\]
And the total number of Queen Cards in a deck = \[4\]
Cards that can be either a heart or a queen is
\[16\]
Cards that are neither a heart nor a queen is
\[52 - 16 = 36\]
Let us assume that \[{\rm{A}}\] be the event that a card drawn is a heart.
So, we can write that a heart card can be drawn from 13 heart cards in \[^{13}{C_1}\] ways
\[{\rm{n}}({\rm{A}}){ = ^{13}}{{\rm{C}}_1}\]
Now, we have to use probability formula, we have
\[P(A) = \dfrac{{n(A)}}{{n(S)}} = \dfrac{{^{13}{C_1}}}{{52}}\]
Now, we have to substitute the corresponding values in the above formula, we get
\[ = \dfrac{{13}}{{52}}\]
Now, let us assume that \[B\] be the event that a Queen card is drawn:
Therefore, a queen card can be drawn from 4 queen cards in \[^4{C_1}\] ways
\[{\rm{n}}(B){ = ^4}{C_1}\]
Now, we have to use probability formula, we have
\[\therefore P(B) = \dfrac{{n(B)}}{{n(S)}} = \dfrac{{^4{C_1}}}{{52}}\]
We know that the total number of Queen Cards in a deck = \[4\]
On substituting the corresponding values from the given data, we get
\[ = \dfrac{4}{{52}}\]
Now, we have to determine the probability of getting queen of heart:
So, we can write it as
\[{\rm{n}}({\rm{A}} \cap {\rm{B}}){ = ^1}{C_1}\]
Therefore, by using probability of Intersection formula
 \[\therefore {\rm{P}}({\rm{A}} \cap {\rm{B}}) = \dfrac{{{\rm{n}}({\rm{A}} \cap {\rm{B}})}}{{{\rm{n}}({\rm{S}})}}\]
Now, on substituting the values, we get
\[ = \dfrac{{^1{{\rm{C}}_1}}}{{52}} = \dfrac{1}{{52}}\]
The needed probability of acquiring a heart or a queen can be determined by\[{\rm{P}}({\rm{ card is a heart or a queen) }}\]
That is, we have to determine
\[P(Q \cup H)\]
Since, we already known that
\[{\rm{P}}({\rm{A}} \cup {\rm{B}}) = {\rm{P}}({\rm{A}}) + {\rm{P}}({\rm{B}}) - {\rm{P}}({\rm{A}} \cap {\rm{B}})\]
Now, let’s substitute the values in the above formula, we have
\[ = \dfrac{{13}}{{52}} + \dfrac{4}{{52}} - \dfrac{1}{{52}}\]
Now, let us have the common denominator as all the terms have same denominator, we get\[ = \dfrac{{13 + 4 - 1}}{{52}}\]
On adding/ subtracting the terms in the numerator, we get
\[ = \dfrac{{16}}{{52}}\]
On further simplification, we get
\[ = \dfrac{4}{{13}}\]
Therefore, the probability that the card will be a queen or a heart is
\[{\rm{P}}({\rm{A}} \cup {\rm{B}}) = \dfrac{4}{{13}}\]



Option ‘C’ is correct



Note: Students should keep in mind that the probability always falls within the range of 0 and 1, inclusive. An event's probability can never be negative or higher than one. So, this concept should be remembered and applied correctly in order to get the desired answer.