
One card is drawn from the pack of $52$ cards randomly. Then find the probability of the card that is to be a king or spade.
A. $\dfrac{1}{26}$
B. $\dfrac{3}{26}$
C. $\dfrac{4}{13}$
D. $\dfrac{3}{13}$
Answer
161.1k+ views
Hint: In this question, we are to find the probability of drawing a card randomly from a pack of cards. For this question, the addition theorem on probability is used. All the given values are substituted in the addition theorem of probability to find the required probability
Formula used: A probability is the ratio of favorable outcomes of an event to the total number of outcomes. So, the probability lies in between \[0\] and \[1\].
The probability is calculated by,
$P(E)=\dfrac{n(E)}{n(S)}$
$n(E)$ - favourable outcomes and $n(S)$ - sample.
If there are two events in a sample space, then the addition theorem on probability is given by
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
When two events happen independently, the occurrence of one is not impacted by the occurrence of the other.
For the events $A$ and $B$, $P(A\cap B)=P(A)\cdot P(B)$ if they are independent and $P(A\cap B)=\Phi $ if they are mutually exclusive.
Complete step by step solution: A pack of cards contains 52 cards.
There are 13 cards of each – Diamonds, Spades, Clubs, and Hearts
There are 4 cards of each – King, Queen, Aces
It is given that; one card is drawn randomly from the pack.
So, the probability of drawing a king card is
$P(K)=\dfrac{4}{52}=\dfrac{1}{13}$
The probability of drawing a spade is
$P(S)=\dfrac{13}{52}=\dfrac{1}{4}$
And the probability of drawing a card that is both a king and a spade is
$P(K\cap S)=\dfrac{1}{52}$
Then, the probability that the card drawn is a king or a spade is
$P(K\text{ }or\text{ }S)=P(K\cup S)$
By the addition theorem on probability,
$\begin{align}
& P(K\cup S)=P(K)+P(S)-P(K\cap S) \\
& \text{ }=\dfrac{1}{13}+\dfrac{1}{4}-\dfrac{1}{52} \\
& \text{ }=\dfrac{4+13-1}{52} \\
& \text{ }=\dfrac{16}{52}=\dfrac{4}{13} \\
\end{align}$
Thus, Option (C) is correct.
Note: In this question, we are to find the probability of an event. For this question, the addition theorem on probability is used. All the given values are substituted in the addition theorem of probability to find the required probability.
Formula used: A probability is the ratio of favorable outcomes of an event to the total number of outcomes. So, the probability lies in between \[0\] and \[1\].
The probability is calculated by,
$P(E)=\dfrac{n(E)}{n(S)}$
$n(E)$ - favourable outcomes and $n(S)$ - sample.
If there are two events in a sample space, then the addition theorem on probability is given by
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
When two events happen independently, the occurrence of one is not impacted by the occurrence of the other.
For the events $A$ and $B$, $P(A\cap B)=P(A)\cdot P(B)$ if they are independent and $P(A\cap B)=\Phi $ if they are mutually exclusive.
Complete step by step solution: A pack of cards contains 52 cards.
There are 13 cards of each – Diamonds, Spades, Clubs, and Hearts
There are 4 cards of each – King, Queen, Aces
It is given that; one card is drawn randomly from the pack.
So, the probability of drawing a king card is
$P(K)=\dfrac{4}{52}=\dfrac{1}{13}$
The probability of drawing a spade is
$P(S)=\dfrac{13}{52}=\dfrac{1}{4}$
And the probability of drawing a card that is both a king and a spade is
$P(K\cap S)=\dfrac{1}{52}$
Then, the probability that the card drawn is a king or a spade is
$P(K\text{ }or\text{ }S)=P(K\cup S)$
By the addition theorem on probability,
$\begin{align}
& P(K\cup S)=P(K)+P(S)-P(K\cap S) \\
& \text{ }=\dfrac{1}{13}+\dfrac{1}{4}-\dfrac{1}{52} \\
& \text{ }=\dfrac{4+13-1}{52} \\
& \text{ }=\dfrac{16}{52}=\dfrac{4}{13} \\
\end{align}$
Thus, Option (C) is correct.
Note: In this question, we are to find the probability of an event. For this question, the addition theorem on probability is used. All the given values are substituted in the addition theorem of probability to find the required probability.
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