Question
Question: From a normal pack of \[52\] cards, one card is selected at random. How can I find the probability o...
From a normal pack of 52 cards, one card is selected at random. How can I find the probability of the card being either a black card or an ace?
Solution
To solve this question, we will first consider two events, one of drawing an ace card and another of drawing a black card. Then, we will find the probability of these events individually. Then we will find the probability of the union of these events using the simple formula:
P(A∪B)=P(A)+P(B)−P(A∩B)
Complete answer:
Let, A be the event of drawing an ace and B be the event of drawing a black coloured card.
So, what we have to find is P(A∪B).
Now we know that there are four aces in a pack. So, one can be chosen in four ways.
Also, the total cards are 52.
∴P(A)=524
⇒P(A)=131
Now we know that out of 52 cards there are a total 26 black cards. So, the probability of event B is;
⇒P(B)=5226
⇒P(B)=21
Now to find P(A∪B), we must be knowing P(A∩B).
Now P(A∩B) means an ace card that is black in colour. Now, from the total card there are two black-coloured aces. So,
P(A∩B)=522
⇒P(A∩B)=261
Now we will put these values in the formula;
P(A∪B)=P(A)+P(B)−P(A∩B)
So, we have;
⇒P(A∪B)=131+21−261
On taking the LCM we have;
⇒P(A∪B)=262+13−1
On solving we get;
⇒P(A∪B)=2614
On reducing the RHS we get;
⇒P(A∪B)=137
Hence the required probability is 137.
Note: One important thing to note is that the probability of an event lies between zero and one. It cannot be more than one. If an event has probability one then, it means that it is a certain event. The probability of non-occurrence of an event is equal to one minus the probability of its occurrence.