13. Probability
Exactly vs at-least, complement trick. Without replacement needs actual counts, not just ratios.
Core ideas
- "At least one" almost always means complement: P(at least one) = 1 - P(none).
- "Exactly k" needs the arrangement count: P(exactly k successes) = (ways to place the successes) * (probability of one specific arrangement).
- Without replacement changes the counts on every draw: update both numerator and denominator each pull.
- Probability = favorable / total only when outcomes are equally likely; otherwise multiply along the branch.
Worked example 1
A bag holds 4 red and 6 blue marbles. Two are drawn without replacement. What is the probability both are red?
Show solution
First draw: 4 red out of 10 marbles, so 4/10. After removing a red, 3 red remain out of 9, so the second draw is 3/9. Multiply: (4/10) * (3/9) = 12/90 = 2/15. The denominators shrink because nothing goes back in.
Worked example 2
A fair coin is flipped 4 times. What is the probability of getting at least one head, and separately, exactly two heads?
Show solution
At least one head: use the complement. P(no heads) = (1/2)^4 = 1/16, so P(at least one) = 1 - 1/16 = 15/16. Exactly two heads: any specific sequence has probability (1/2)^4 = 1/16, and the two heads can land in 4C2 = 6 positions. So P(exactly two) = 6/16 = 3/8. Note the split: at-least used complement, exactly used a placement count.
Practice set
Question 1
A committee of 2 people is selected at random from a group of 5 men and 3 women. What is the probability that the committee contains exactly one woman?
Exactly one woman means one woman and one man; count those pairs against all possible pairs.
The number of ways to pick 2 people from 8 is 8 choose 2, which is 28. Committees with exactly one woman pair one of 3 women with one of 5 men, giving 3 times 5 = 15 committees. The probability is 15/28. The trap 3/28 counts two women instead of exactly one.
Question 2
A jar contains 8 blue marbles and r red marbles. If two marbles are drawn at random without replacement, the probability that both are blue is 14/33. What is the value of r?
Write the two-draw probability in terms of the total count n and solve n(n-1).
Let n be the total number of marbles. The probability of two blues without replacement is (8/n)(7/(n-1)) = 56/(n(n-1)). Setting 56/(n(n-1)) = 14/33 gives n(n-1) = 56 times 33/14 = 132, so n = 12. Then r = 12 - 8 = 4. Picking 12 answers the wrong question, the total rather than the number of red marbles.
Question 3
A fair six-sided die is rolled twice. What is the probability that at least one of the rolls shows a 6?
Adding 1/6 and 1/6 double counts something; try the no-sixes complement instead.
The probability that a single roll is not a 6 is 5/6, so the probability that neither roll is a 6 is (5/6)(5/6) = 25/36. The complement, at least one 6, is 1 - 25/36 = 11/36. The answer 1/3 comes from adding 1/6 + 1/6, which double counts the outcome where both rolls are 6.
Question 4
A bag contains 4 red marbles and 6 blue marbles. If two marbles are drawn at random without replacement, what is the probability that both are blue?
The second draw comes from a smaller bag, so use the updated counts, not the original ratio.
The first blue is drawn with probability 6/10. With one blue removed, the second blue is drawn with probability 5/9. Multiplying gives (6/10)(5/9) = 30/90 = 1/3. Choosing 9/25 comes from wrongly using 6/10 twice, as if the marble were replaced.