
One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is (a) an ace (b) red (c) either red or king (d) red and a king (e) a face card (f) a red face card (g) “2” of spades (h) “10” of a black suit.
Answer
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Hint: We need to find the probability of the given events. We will be solving the given question by finding out all the favorable outcomes and all the possible outcomes for every event. Then, we find the ratio of the favorable outcomes to the total outcomes to get the desired probability.
Complete step by step answer:
We are asked to find the probability for the given set of events. We will be solving the given question using the concept of probability.
According to our question,
One card is drawn from the pack of 52 cards.
(a) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is 4.
It is given as follows,
$\Rightarrow n\left( ace \right)=4$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is ace is given as follows,
$\Rightarrow P\left( ace \right)=\dfrac{n\left( ace \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( ace \right)=\dfrac{4}{52}$
Canceling out the common factors, we get,
$\therefore P\left( ace \right)=\dfrac{1}{13}$
(b) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is 26.
It is given as follows,
$\Rightarrow n\left( red \right)=26$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is red is given as follows,
$\Rightarrow P\left( red \right)=\dfrac{n\left( red \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( red \right)=\dfrac{26}{52}$
Canceling out the common factors, we get,
$\therefore P\left( red \right)=\dfrac{1}{2}$
(c) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is given as follows,
$\Rightarrow n\left( red \right)=26$
$\Rightarrow n\left( king \right)=2$
$\Rightarrow n\left( red\text{ or }king \right)=n\left( red \right)+n\left( king \right)$
Substituting the values, we get,
$\Rightarrow n\left( red\text{ or }king \right)=26+2$
$\Rightarrow n\left( red\text{ or }king \right)=28$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is red or the king is given as follows,
$\Rightarrow P\left( red\text{ or }king \right)=\dfrac{n\left( red\text{ or }king \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( red\text{ or }king \right)=\dfrac{28}{52}$
Canceling out the common factors, we get,
$\therefore P\left( red\text{ or }king \right)=\dfrac{7}{13}$
(d) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is given as follows,
$\Rightarrow n\left( red \right)=26$
$\Rightarrow n\left( king \right)=2$
$\Rightarrow n\left( red\text{ and }king \right)=n\left( red \right)\cap n\left( king \right)$
From the above, we get,
$\Rightarrow n\left( red\text{ and }king \right)=2$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is red and a king is given as follows,
$\Rightarrow P\left( red\text{ and }king \right)=\dfrac{n\left( red\text{ and }king \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( red\text{ and }king \right)=\dfrac{2}{52}$
Canceling out the common factors, we get,
$\therefore P\left( red\text{ and }king \right)=\dfrac{1}{26}$
(e) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is 12.
It is given as follows,
$\Rightarrow n\left( face \right)=12$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is a face card is given as follows,
$\Rightarrow P\left( face \right)=\dfrac{n\left( face \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( face \right)=\dfrac{12}{52}$
Canceling out the common factors, we get,
$\therefore P\left( face \right)=\dfrac{3}{13}$
(f) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is given as follows,
$\Rightarrow n\left( red\text{ }face \right)=6$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is a red face card is given as follows,
$\Rightarrow P\left( red\text{ }face \right)=\dfrac{n\left( red\text{ }face \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( red\text{ }face \right)=\dfrac{6}{52}$
Canceling out the common factors, we get,
$\therefore P\left( red\text{ }face \right)=\dfrac{3}{26}$
(g) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is given as follows,
$\Rightarrow n\left( 2\text{ of spades} \right)=1$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is a red face card is given as follows,
$\Rightarrow P\left( 2\text{ of spades} \right)=\dfrac{n\left( 2\text{ of spades} \right)}{n\left( S \right)}$
Substituting the values, we get,
$\therefore P\left( 2\text{ of spades} \right)=\dfrac{1}{52}$
(h) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is given as follows,
$\Rightarrow n\left( \text{10 of black} \right)=2$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is 10 of black suit is given as follows,
$\Rightarrow P\left( \text{10 of black} \right)=\dfrac{n\left( \text{10 of black} \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( \text{10 of black} \right)=\dfrac{2}{52}$
Canceling out the common factors, we get,
$\therefore P\left( \text{10 of black} \right)=\dfrac{1}{26}$
Note: The given question is directly formula based and any mistakes in writing the formula of probability will result in an incorrect solution. The events that have the same probability of occurring are called equally likely events.
Complete step by step answer:
We are asked to find the probability for the given set of events. We will be solving the given question using the concept of probability.
According to our question,
One card is drawn from the pack of 52 cards.
(a) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is 4.
It is given as follows,
$\Rightarrow n\left( ace \right)=4$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is ace is given as follows,
$\Rightarrow P\left( ace \right)=\dfrac{n\left( ace \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( ace \right)=\dfrac{4}{52}$
Canceling out the common factors, we get,
$\therefore P\left( ace \right)=\dfrac{1}{13}$
(b) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is 26.
It is given as follows,
$\Rightarrow n\left( red \right)=26$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is red is given as follows,
$\Rightarrow P\left( red \right)=\dfrac{n\left( red \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( red \right)=\dfrac{26}{52}$
Canceling out the common factors, we get,
$\therefore P\left( red \right)=\dfrac{1}{2}$
(c) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is given as follows,
$\Rightarrow n\left( red \right)=26$
$\Rightarrow n\left( king \right)=2$
$\Rightarrow n\left( red\text{ or }king \right)=n\left( red \right)+n\left( king \right)$
Substituting the values, we get,
$\Rightarrow n\left( red\text{ or }king \right)=26+2$
$\Rightarrow n\left( red\text{ or }king \right)=28$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is red or the king is given as follows,
$\Rightarrow P\left( red\text{ or }king \right)=\dfrac{n\left( red\text{ or }king \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( red\text{ or }king \right)=\dfrac{28}{52}$
Canceling out the common factors, we get,
$\therefore P\left( red\text{ or }king \right)=\dfrac{7}{13}$
(d) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is given as follows,
$\Rightarrow n\left( red \right)=26$
$\Rightarrow n\left( king \right)=2$
$\Rightarrow n\left( red\text{ and }king \right)=n\left( red \right)\cap n\left( king \right)$
From the above, we get,
$\Rightarrow n\left( red\text{ and }king \right)=2$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is red and a king is given as follows,
$\Rightarrow P\left( red\text{ and }king \right)=\dfrac{n\left( red\text{ and }king \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( red\text{ and }king \right)=\dfrac{2}{52}$
Canceling out the common factors, we get,
$\therefore P\left( red\text{ and }king \right)=\dfrac{1}{26}$
(e) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is 12.
It is given as follows,
$\Rightarrow n\left( face \right)=12$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is a face card is given as follows,
$\Rightarrow P\left( face \right)=\dfrac{n\left( face \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( face \right)=\dfrac{12}{52}$
Canceling out the common factors, we get,
$\therefore P\left( face \right)=\dfrac{3}{13}$
(f) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is given as follows,
$\Rightarrow n\left( red\text{ }face \right)=6$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is a red face card is given as follows,
$\Rightarrow P\left( red\text{ }face \right)=\dfrac{n\left( red\text{ }face \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( red\text{ }face \right)=\dfrac{6}{52}$
Canceling out the common factors, we get,
$\therefore P\left( red\text{ }face \right)=\dfrac{3}{26}$
(g) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is given as follows,
$\Rightarrow n\left( 2\text{ of spades} \right)=1$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is a red face card is given as follows,
$\Rightarrow P\left( 2\text{ of spades} \right)=\dfrac{n\left( 2\text{ of spades} \right)}{n\left( S \right)}$
Substituting the values, we get,
$\therefore P\left( 2\text{ of spades} \right)=\dfrac{1}{52}$
(h) In this case, the number of total outcomes is 52.
It is given as follows,
$\Rightarrow n\left( S \right)=52$
The number of favorable outcomes is given as follows,
$\Rightarrow n\left( \text{10 of black} \right)=2$
We know that
$\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}$
Following the same, the probability that the card drawn is 10 of black suit is given as follows,
$\Rightarrow P\left( \text{10 of black} \right)=\dfrac{n\left( \text{10 of black} \right)}{n\left( S \right)}$
Substituting the values, we get,
$\Rightarrow P\left( \text{10 of black} \right)=\dfrac{2}{52}$
Canceling out the common factors, we get,
$\therefore P\left( \text{10 of black} \right)=\dfrac{1}{26}$
Note: The given question is directly formula based and any mistakes in writing the formula of probability will result in an incorrect solution. The events that have the same probability of occurring are called equally likely events.
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