1. Suppose I roll four 10-sided dice.
b. What is the probability that I get at least one match (that means I get the same number at least twice)
the opposite of at least one match is all different. If I don't get all different then I get at least one match (and vice versa)
P(at least one match) = 1 - P(all different)
1 - P(first die can be anything)*P(second die not same as first) * P(third not same as either of first two) * P(fourth not same as any of first three)
The first die can have any number and still have all four be different. There are 10 sides on the die, and 10 are allowed: P=10/10=1
The second die has to have a differnt number from the first one in order for all to be different. If the first die showed 4, then the second could show any number but 4 and be different. 9 of the 10 sides are different: P=9/10
The third die must avoid the two numbers showing on the first two dice, so 8 of the 10 possible numbers will give something different. P=8/10
The fourth die must avoid the 3 numbers showing on the first 3 dice, so 7 of the 10 possible numbers will give sumething different. P=7/10
1 - (10/10)(9/10)(8/10)(7/10)
Multiply before adding: multiply all of the numerators together to get the numerator, and all of the denominators to get the denominator.
1 - 5040/10000
1 - 63/125 = 62/125 = .496
3. I have a regular deck of cards, and I draw 3 without replacement
a. What is the probability that I get at least two of a kind (same number or face card)
the opposite of at least two of a kind is all different kinds. If I don't get all different kinds then I get at least two of the same kind (and vice versa)
P(at least two of the same kind) = 1 - P(all different kinds)
1 - P(first is any card) * P(second not same kind as first) * P(third not same kind as previous two)
To begin with, you can pick any card: all 52 of the 52 cards will make it possible to have all 3 different kinds
There are 13 kinds of cards: A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. You have picked one. let's pretend it is a J. Then there are 4A's, 4 2'3, 4 3's, ..., 4 10's, 3 J's, 4 Q's, 4 K's left. To be different you can pick any of A, 2, 3, 4, 5, 6, 7, 8, 9, 10, Q, K. There are 48 of these cards, and 51 total cards. P = 48/51
Now you have picked two different kinds of cards. Let's pretend you have picked a J and a10. To be different you can pick an A, 2, 3, 4, 5, 6, 7, 8, 9, Q, K. There are 4 each of these left for a total of 44, out of 50 cards left P=44/50
1 - (52/52)(48/51)(44/50)
1 - 109824/132600
1 - 352/425 = 73/425 = .172