Question

Ten red cards and ten black cards are placed in a bag. You choose one card and then another without replacing the first card. What is the probability

Probability and combinatorics
Ten red cards and ten black cards are placed in a bag. You choose one card and then another without replacing the first card. What is the probability that the first card will be red and the second card will be black?

2021-06-01

The events are dependent because the outcome of drawing the first card affects the outcome of the next draw. With two dependent events A and B, we also apply the Multiplication Rule:
$$P(A\ then\ B) = P(A) - P(B\ after\ A)$$
The probability that the first draw is a red card is $$38 = 3$$. The probability that the second draw is a black card is now 48 (since there is no replacement). Hence, the probability of first drawing red card then a black is:
$$P( \text{red then black}) = P(red) - P(\text{black after red})$$
(ved thon black) = $$\displaystyle\frac{{1}}{{2}}\cdot\frac{{10}}{{19}}$$
P(red then black) = $$\displaystyle\frac{{5}}{{19}}$$