
A card is drawn from a well-shuffled pack of $52$ cards. Events A and B are defined as follows
A: Getting a card of spade
B: Getting an ace, then A and B are
Answer
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Hint: First, a pack of cards means it will contain $52$ cards.
In four different shapes, they are $13$ cards each thus totals fifty-two cards in a pack of cards.
The shapes are heart, spade, diamond, and clubs.
Each will contain thirteen cards.
Formula used: \[{\text{Probability}} = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\]
Complete step-by-step solution:
Since in a pack of cards there are fifty-two cards, also in each of the four shapes there are thirteen cards like $1,2,3,...,10$ and jack, king and queen.
A: Getting a card of spade
There are a total of thirteen spade cards in the given set of cards, thus we get spade cards$ = 13$.
Since the \\text{Probability} = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\],
Here in event A, the number of favorable events is spade cards$ = 13$ and the number of total events is cards containing $52$cards.
Thus, we get the probability of getting the spade card is \[\text{Probability} = \dfrac{{13}}{{52}}\].
Further simplifying we get \[\text{Probability} = \dfrac{1}{4}\] is the probability of getting the spade cards.
B: Getting an ace
Now we are going to find the probability of getting the ace cards since there are only four ace cards in the shapes: heart, diamond, spade, and clubs.
Ace is also called the number one in the card game.
Thus, we get \[\text{Probability}= \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\], where several favorable events are four and total events are fifty-two.
Hence, we get \[\text{Probability} = \dfrac{4}{{52}} \Rightarrow \dfrac{1}{{13}}\] which is the probability of getting the card ace.
A and B are
Hence event A and event B are independent.
The density of the spade cards does not equal the density of the card's ace.
Hence the required result is done
Note: Since \[\text{Probability} = \dfrac{1}{4}\] is the chance of the spade; which means in four cards there is a possibility of one spade card.
Also, the \[\text{Probability} \Rightarrow \dfrac{1}{{13}}\] is the chance of getting the card ace, which means in the thirteen cards there is a possibility of getting the card ace.
In four different shapes, they are $13$ cards each thus totals fifty-two cards in a pack of cards.
The shapes are heart, spade, diamond, and clubs.
Each will contain thirteen cards.
Formula used: \[{\text{Probability}} = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\]
Complete step-by-step solution:
Since in a pack of cards there are fifty-two cards, also in each of the four shapes there are thirteen cards like $1,2,3,...,10$ and jack, king and queen.
A: Getting a card of spade
There are a total of thirteen spade cards in the given set of cards, thus we get spade cards$ = 13$.
Since the \\text{Probability} = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\],
Here in event A, the number of favorable events is spade cards$ = 13$ and the number of total events is cards containing $52$cards.
Thus, we get the probability of getting the spade card is \[\text{Probability} = \dfrac{{13}}{{52}}\].
Further simplifying we get \[\text{Probability} = \dfrac{1}{4}\] is the probability of getting the spade cards.
B: Getting an ace
Now we are going to find the probability of getting the ace cards since there are only four ace cards in the shapes: heart, diamond, spade, and clubs.
Ace is also called the number one in the card game.
Thus, we get \[\text{Probability}= \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\], where several favorable events are four and total events are fifty-two.
Hence, we get \[\text{Probability} = \dfrac{4}{{52}} \Rightarrow \dfrac{1}{{13}}\] which is the probability of getting the card ace.
A and B are
Hence event A and event B are independent.
The density of the spade cards does not equal the density of the card's ace.
Hence the required result is done
Note: Since \[\text{Probability} = \dfrac{1}{4}\] is the chance of the spade; which means in four cards there is a possibility of one spade card.
Also, the \[\text{Probability} \Rightarrow \dfrac{1}{{13}}\] is the chance of getting the card ace, which means in the thirteen cards there is a possibility of getting the card ace.
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