# A card is drawn from a standard deck, its color is noted, and then it is replaced in the deck. Another card is then drawn from the deck. What is the probability that the first card is red and the second card is black?

##### 1 Answer
Jan 11, 2015

The probability is $\left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) = \frac{1}{4}$

The Probability of the first card being red is $\frac{26}{52} = \frac{1}{2}$

Since you put the card back, the probability of the second card being black is again $\frac{26}{52} = \frac{1}{2}$

The two events are independent: the second draw starts at the same "square one" as the first. In this case we can say:
Event one AND event two, which means we can multiply the chances:

1/2*1/2=1/4=0.25=25%

This goes for any sequence. The other ones being red-red, black-red, and black-black. Together they make up 4 sequences, with a total probability of 4/4=1=4*25%=100%. Nice, isn't it?

Extra:
If you didn't put the first card back, and if it were a red card, then the probability of getting a black card on the second pull would be slightly higher: 26/51~~0.51=51% instead of 50%