
If a card is drawn from a pack of cards. The probability of getting black ace is
A) \[\dfrac{1}{{52}}\]
B) \[\dfrac{1}{{26}}\]
C) \[\dfrac{1}{{13}}\]
D) \[\dfrac{1}{4}\]
Answer
558.9k+ views
Hint:
Here, we will use the fact a pack of cards contains 52 cards. We will find the favorable outcomes of getting a black ace when a card is drawn from these 52 cards. Substituting the favorable outcomes and the total number of outcomes in the formula of probability, we will be able to find the required probability of getting black ace.
Formula Used:
Probability, \[P = \]Favorable outcomes\[ \div \]Total number of outcomes
Complete step by step solution:
We know that, in a pack of cards, there are a total 52 cards.
Thus, the total number of possible cases\[ = 52\]
Now, these 52 cards are divided into red and black cards consisting of 26 cards each.
Further, the red cards are divided into red hearts and red diamonds which contain 13 cards each.
Similarly, the black cards are divided into black club and black spade which contain 13 cards each.
These 13 cards are: \[A,2,3,4,....,10,J,Q,K\]
Now, according to the question, we are required to find the probability of getting a black ace.
We know that there are 26 black cards in which we have 1 ace of club and 1 ace of spade.
Thus, the total possible ways of getting a black ace\[ = 2\]
Hence, the number of favorable outcomes\[ = 2\]
Now using the formula of Probability, \[P = \]Favorable outcomes\[ \div \]Total number of outcomes, we get
The probability of getting black ace\[ = \dfrac{2}{{52}} = \dfrac{1}{{26}}\]
Therefore, the required probability of getting a black ace is \[\dfrac{1}{{26}}\]
Hence, option B is the correct answer.
Note:
Probability tells us the extent to which an event is likely to occur, i.e. the possibility of the occurrence of an event. Hence, it is measured by dividing the favorable outcomes by the total number of outcomes. The probability of an event can never be greater than 1 or less than 0. The probability of a sure event is 1 and the probability of an impossible event is 0. We use probability in daily life to predict the total score in a cricket match, weather forecasting, etc.
Here, we will use the fact a pack of cards contains 52 cards. We will find the favorable outcomes of getting a black ace when a card is drawn from these 52 cards. Substituting the favorable outcomes and the total number of outcomes in the formula of probability, we will be able to find the required probability of getting black ace.
Formula Used:
Probability, \[P = \]Favorable outcomes\[ \div \]Total number of outcomes
Complete step by step solution:
We know that, in a pack of cards, there are a total 52 cards.
Thus, the total number of possible cases\[ = 52\]
Now, these 52 cards are divided into red and black cards consisting of 26 cards each.
Further, the red cards are divided into red hearts and red diamonds which contain 13 cards each.
Similarly, the black cards are divided into black club and black spade which contain 13 cards each.
These 13 cards are: \[A,2,3,4,....,10,J,Q,K\]
Now, according to the question, we are required to find the probability of getting a black ace.
We know that there are 26 black cards in which we have 1 ace of club and 1 ace of spade.
Thus, the total possible ways of getting a black ace\[ = 2\]
Hence, the number of favorable outcomes\[ = 2\]
Now using the formula of Probability, \[P = \]Favorable outcomes\[ \div \]Total number of outcomes, we get
The probability of getting black ace\[ = \dfrac{2}{{52}} = \dfrac{1}{{26}}\]
Therefore, the required probability of getting a black ace is \[\dfrac{1}{{26}}\]
Hence, option B is the correct answer.
Note:
Probability tells us the extent to which an event is likely to occur, i.e. the possibility of the occurrence of an event. Hence, it is measured by dividing the favorable outcomes by the total number of outcomes. The probability of an event can never be greater than 1 or less than 0. The probability of a sure event is 1 and the probability of an impossible event is 0. We use probability in daily life to predict the total score in a cricket match, weather forecasting, etc.
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