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Question: One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Fi...

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.

Explanation

Solution

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,
⇒n(S)=52\Rightarrow n\left( S \right)=52
The number of favorable outcomes is 4.
It is given as follows,
⇒n(ace)=4\Rightarrow n\left( ace \right)=4
We know that
⇒probability=favorable outcomestotal outcomes\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}
Following the same, the probability that the card drawn is ace is given as follows,
⇒P(ace)=n(ace)n(S)\Rightarrow P\left( ace \right)=\dfrac{n\left( ace \right)}{n\left( S \right)}
Substituting the values, we get,
⇒P(ace)=452\Rightarrow P\left( ace \right)=\dfrac{4}{52}
Canceling out the common factors, we get,
∴P(ace)=113\therefore P\left( ace \right)=\dfrac{1}{13}
(b) In this case, the number of total outcomes is 52.
It is given as follows,
⇒n(S)=52\Rightarrow n\left( S \right)=52
The number of favorable outcomes is 26.
It is given as follows,
⇒n(red)=26\Rightarrow n\left( red \right)=26
We know that
⇒probability=favorable outcomestotal outcomes\Rightarrow probability=\dfrac{favorable\text{ }outcomes}{total\text{ }outcomes}
Following the same, the probability that the card drawn is red is given as follows,
⇒P(red)=n(red)n(S)\Rightarrow P\left( red \right)=\dfrac{n\left( red \right)}{n\left( S \right)}
Substituting the values, we get,
⇒P(red)=2652\Rightarrow P\left( red \right)=\dfrac{26}{52}
Canceling out the common factors, we get,
∴P(red)=12\therefore P\left( red \right)=\dfrac{1}{2}
(c) In this case, the number of total outcomes is 52.
It is given as follows,
⇒n(S)=52\Rightarrow n\left( S \right)=52
The number of favorable outcomes is given as follows,
⇒n(red)=26\Rightarrow n\left( red \right)=26
⇒n(king)=2\Rightarrow n\left( king \right)=2
⇒n(red or king)=n(red)+n(king)\Rightarrow n\left( red\text{ or }king \right)=n\left( red \right)+n\left( king \right)
Substituting the values, we get,
⇒n(red or king)=26+2\Rightarrow n\left( red\text{ or }king \right)=26+2
⇒n(red or king)=28\Rightarrow n\left( red\text{ or }king \right)=28
We know that
⇒probability=favorable outcomestotal outcomes\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,
⇒P(red or king)=n(red or king)n(S)\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,
⇒P(red or king)=2852\Rightarrow P\left( red\text{ or }king \right)=\dfrac{28}{52}
Canceling out the common factors, we get,
∴P(red or king)=713\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,
⇒n(S)=52\Rightarrow n\left( S \right)=52
The number of favorable outcomes is given as follows,
⇒n(red)=26\Rightarrow n\left( red \right)=26
⇒n(king)=2\Rightarrow n\left( king \right)=2
⇒n(red and king)=n(red)∩n(king)\Rightarrow n\left( red\text{ and }king \right)=n\left( red \right)\cap n\left( king \right)
From the above, we get,
⇒n(red and king)=2\Rightarrow n\left( red\text{ and }king \right)=2
We know that
⇒probability=favorable outcomestotal outcomes\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,
⇒P(red and king)=n(red and king)n(S)\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,
⇒P(red and king)=252\Rightarrow P\left( red\text{ and }king \right)=\dfrac{2}{52}
Canceling out the common factors, we get,
∴P(red and king)=126\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,
⇒n(S)=52\Rightarrow n\left( S \right)=52
The number of favorable outcomes is 12.
It is given as follows,
⇒n(face)=12\Rightarrow n\left( face \right)=12
We know that
⇒probability=favorable outcomestotal outcomes\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,
⇒P(face)=n(face)n(S)\Rightarrow P\left( face \right)=\dfrac{n\left( face \right)}{n\left( S \right)}
Substituting the values, we get,
⇒P(face)=1252\Rightarrow P\left( face \right)=\dfrac{12}{52}
Canceling out the common factors, we get,
∴P(face)=313\therefore P\left( face \right)=\dfrac{3}{13}
(f) In this case, the number of total outcomes is 52.
It is given as follows,
⇒n(S)=52\Rightarrow n\left( S \right)=52
The number of favorable outcomes is given as follows,
⇒n(red face)=6\Rightarrow n\left( red\text{ }face \right)=6
We know that
⇒probability=favorable outcomestotal outcomes\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,
⇒P(red face)=n(red face)n(S)\Rightarrow P\left( red\text{ }face \right)=\dfrac{n\left( red\text{ }face \right)}{n\left( S \right)}
Substituting the values, we get,
⇒P(red face)=652\Rightarrow P\left( red\text{ }face \right)=\dfrac{6}{52}
Canceling out the common factors, we get,
∴P(red face)=326\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,
⇒n(S)=52\Rightarrow n\left( S \right)=52
The number of favorable outcomes is given as follows,
⇒n(2 of spades)=1\Rightarrow n\left( 2\text{ of spades} \right)=1
We know that
⇒probability=favorable outcomestotal outcomes\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,
⇒P(2 of spades)=n(2 of spades)n(S)\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,
∴P(2 of spades)=152\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,
⇒n(S)=52\Rightarrow n\left( S \right)=52
The number of favorable outcomes is given as follows,
⇒n(10 of black)=2\Rightarrow n\left( \text{10 of black} \right)=2
We know that
⇒probability=favorable outcomestotal outcomes\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,
⇒P(10 of black)=n(10 of black)n(S)\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,
⇒P(10 of black)=252\Rightarrow P\left( \text{10 of black} \right)=\dfrac{2}{52}
Canceling out the common factors, we get,
∴P(10 of black)=126\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.