Question

A card is drawn from a pack of 52 cards. Find the probability that the drawn card is
A. Neither a diamond nor a face card.
B. King or red card.
C. Black and face card.

Answer 1

Hint: First we should count the total number of favourable outcomes which is 2 and use it as $ n\left( E \right) $. Then count the event that can occur possibly in the sample space, which is 52. Then we will use the formula $ P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)} $ , where $ P\left( E \right) $ is the possibility of the event that is asked, $ n\left( E \right) $ is the number of favourable outcomes and $ n\left( S \right) $ is the number of all the events that can possibly occur.

Complete step-by-step answer:
In the question we are given that a pack of cards is there and a card is drawn out of a pack of 52 cards. We have been asked to find out the probability of finding neither a diamond nor a face card, king or red card, black and face card. We will first define what probability is and understand the basic terms related to probability to be used in the question. The probability of an event is a measure of the likelihood that the event would occur. If an experiments’ outcomes are equally likely to occur, then the probability of an event E is the number of outcomes in E divided by the number of outcomes in the sample space. Here, sample space consists of all the events that can occur possibly. So, it can be written as, $ P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)} $ , where $ P\left( E \right) $ is the possibility of the event that is asked, $ n\left( E \right) $ is the number of favourable outcomes and $ n\left( S \right) $ is the number of all the events that can possibly occur. Now, we should know a little about the cards too, on which the question is based. In a pack of 52 cards, there are 4 suits available, which are Spade, Heart, Club and Diamond. All have 13 cards each. Each suit has 1 King, 1 Queen, 1 Jack, 1 Ace and 9 cards that are numbered 2 - 10.
So, here, we are asked to find the probability of getting a Queen or King of Heart. The sample space consists of all the 52 cards in the pack. So, $ n\left( S \right)=52 $ . Now, we will take each case individually.
A. Neither a diamond nor a face card.
We know that there are 13 cards of Diamond and a total of $ \left( 3\times 4 \right)=12 $ face cards in a pack of 52 cards. But out of that 3 face cards are Diamond also, so the face cards left would be 9. So, the probability of the possibility of an event would be $ 52-\left( 13+9 \right)=52-22=30 $ . So, we get $ n\left( E \right)=30 $ . So, the probability is,
 $ \begin{align}
  & P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)} \\
 & \Rightarrow P\left( E \right)=\dfrac{30}{52} \\
 & \Rightarrow P\left( E \right)=\dfrac{15}{26} \\
\end{align} $
Hence, the probability of neither a Diamond nor a face card is $ \dfrac{15}{26} $.

B. King or red card.
We know that are $ 13\times 2=26 $ red cards. There are 4 Kings too, but out of it 2 are red, so we have 2 left. Therefore, we will get the possibility of the event being, $ n\left( E \right)=28 $ . So, the probability is,
 $ \begin{align}
  & P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)} \\
 & \Rightarrow P\left( E \right)=\dfrac{28}{52} \\
 & \Rightarrow P\left( E \right)=\dfrac{7}{13} \\
\end{align} $
Hence, the probability of a King or a red card is $ \dfrac{7}{13} $.

C. Black and face card.
We know that there are 3 face cards of each suit and there are 2 sets of black suits, so black and face cards would be $ 3\times 2=6 $ . So, $ n\left( E \right)=6 $ . So, we get the probability as,
 $ \begin{align}
  & P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)} \\
 & \Rightarrow P\left( E \right)=\dfrac{6}{52} \\
 & \Rightarrow P\left( E \right)=\dfrac{3}{26} \\
\end{align} $
Hence, the probability of black and face cards is $ \dfrac{3}{26} $ .

Note: The students should be familiar with packs of cards and the number of suits of each colour and the face cards. They should also not get confused with the face cards and remember that the face cards do not include the Ace in them.