Bold numbers are good final answers
Red words are explanations of the numbers
1. Suppose I roll four 10-sided dice.
a. What is the probability that I roll at least one 9
the opposite of at least one 9 is no 9's. If I don't get no 9's then I get at least one 9 (and vice versa)
P(at least one 9) = 1 - P(no 9's)
1 - P(not a 9)* P(not a 9)* P(not a 9)* P(not a 9)
1 - (9/10)(9/10)(9/10)(9/10) for each die the probability is 9/10 that the number rolled is not a 9.
1 - 6561/10000
10000/10000 - 6561/10000 = 3439/10000 = .3439
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)
1 - (10/10)(9/10)(8/10)(7/10)
1 - 5040/10000
1 - 63/125 = 62/125 = .496
2. I have a bag with 3 red, 3 blue, 3 green and 3 white marbles, and I pick 4 without replacement
a. What is the probability that I get at least one red
the opposite of at least one red is no reds. If I don't get no reds then I get at least one red (and vice versa)
P(at least one red) = 1 - P(no reds)
1 - P(first not red) * P(second not red) * P(third not red) * P(fourth not red)
1 - (9/12)(8/11)(7/10)(6/9)
1 - 3024/11880
1 - 14/55 = 41/55 = .745
b. What is the probability that I get at least two of the same color
the opposite of at least two of the same color is all different colors. If I don't get all different colors then I get at least two of the same color (and vice versa)
P(at least two of the same color) = 1 - P(all different colors)
1 - P(first any color) * P(second different color from first) * P(third different color from first two) * P(fourth different color from any of first three)
1 - (12/12)(9/11)(6/10)(3/9)
1 - 1944/11880
1 - 9/55 = 46/55 = .836
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)
1 - (52/52)(48/51)(44/50)
1 - 109824/132600
1 - 352/425 = 73/425 = .172
b. What is the probability that I get at least one Jack
the opposite of at least one jack is no jacks. If I don't get no jacks then I get at least one jack (and vice versa)
P(at least one jack) = 1 - P(no jacks)
1 - P(not a Jack) * P(not a jack)
1 - (48/52)(47/52)
1 - 2256/2704
1 - 141/169 = 28/169 = .166