Question

The following exercise refers to choosing two cards from a thoroughly shuffled deck. Assume that the deck is shuffled after a card is returned to the deck.
If you do not put the first card back in the deck before you draw the next, what is the probability that the first card is a club and the second one is black? (Enter your probability as a fraction.)

Answer 1

Using it's concept, it is found that there is a $\frac{13}{102}$ probability that the first card is a club and the second one is black.

What is a probability?

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem:

• In a deck, there are 52 cards, of which 13 are clubs, hence $P(A) = \frac{13}{52} = \frac{1}{4}$.

• Then, of the remaining 51 cards, 26 are black, hence $P(B) = \frac{26}{51}$.

Then:

$p = P(A)P(B) = \frac{1}{4} \times \frac{26}{51} = \frac{13}{102}$

$\frac{13}{102}$ probability that the first card is a club and the second one is black.

More can be learned about the probability concept at brainly.com/question/15536019