The question seems a bit erroneous hence I am writing the correct question here and then answering it.
A standard deck of 52 cards contains 4 suits: hearts, clubs, diamonds, and spades. Each suit consists of cards numbered 2 through 10, a jack, a queen, a king, and an ace. Zhi Qing decides to pick one card at random from a standard deck of 52 cards. Let A be the event that she chooses a three card and B be the event that she chooses a seven card. What is P(A or B), the probability that the card Zhi Qing chooses is either a three or a seven?
Answer:
The probability that the card Zhi Qing chooses is either a three or a seven is 8/52.
Explanation:
Event A: She chooses a three card. We can calculate the probability of event A as:
P(A) = No. of 3 numbered cards/Total no. of cards
Since there are 4 suits and each suit has 1 three-numbered card, so there are a total of 4 three-numbered cards.
P(A) = 4/52
Event B = She chooses a seven card. We can calculate the probability as:
P(B) = No. of seven-numbered cards/Total no. of cards
Again, there are 4 suits and each suit has 1 seven-numbered card. So, the number of 7-numbered cards is 4.
P(B) = 4/52
The probability that Zhi Qing chooses either a three or a seven is:
P(A or B) = P(A) + P(B) - P(A and B)
P(A and B) = 0 because a card can not be a seven and a 3 at the same time. These events are mutually exclusive hence P(A and B)is equal to zero. So,
P(A or B) = P(A) + P(B)
= 4/52 + 4/52
P(A or B) = 8/52