Lets say we pick the first card as a Red and 10. Now there are 49 cards remaining out of which 9 cards have the color red and 4 cards have the number 10 i.e. 13 cases that fail our condition of \”not of same color and not of same number\”. Hence probability of success = (49-13)/49 = 36/49

So here’s the way I’m thinking about this one. We’re drawing two cards and we want to make sure they’re not the same number or same color. So the first card we draw can be anything (e.g. lets say, blue and number 5). It’s the second card that we have to worry about…

And for the second card, we’re calculating the P(not blue and not 5). Because these two events aren’t independent, P(not blue and not 5)=P(not blue)* P(not 5 | not blue). P(not blue) is 40/49. P(not 5 | not blue) is 36/40 because there are four 5’s left, and we’re only looking at the subset of cards that are not blue (40 cards). Therefore, the answer should be .734 (36/40 * 40/49)

Source

There are C(50,2)=50*49/2!=25*49=1225 ways to pick 2 cards. If the 1st card is chosen (50 ways), there are 36 cards left that will make up the desired pair, so there are 50*36/2!=25*36 desired outcomes. The P=(25*36)/(25*49)=36/49

The Number of combinations for selecting two cards is (50 C 2) = 25*49— Which goes in Denominator of the Probability

Now to evaluate the number of favorable combinations, we need to select 2 colors from the given 5 colors i.e. (5 C 2), and from each of the 10 color set cards we need to select 1 card each i.e. (10 C 1) combinations, ass all the 3 events are independent the number of favorable combinations = (5 C 2)*(10 C 1)*(10 C 1) = 25*40

Therefore, The probability that we select two cards of different color = (25*40)/(25*49) = 40/49

Prob of getting two different color cards = 40/49 Prob of getting two different number cards = 45/49 Prob of getting different color and number = 40/49 * 45/49 =0.74

The probability of picking two cards with the same color and the same number WITHOUT replacement is 0 because If we pick “red 5”, there is no more “red 5” left in the deck.

With the replacement its a different story:

- We pick the first card, let’s say it is “red 5”. The probability of picking it is 1/50.
- We put it back. shuffle a deck
- We pick the second card. The probability of picking “red 5” again is 1/50
- The probability of NOT picking “red 5” two times is 1 – (1/50 * 1/50) = 0.9996