basic probability

Problem Context: Tickets in Boxes

One ticket will be drawn at random from each of the two boxes shown below:

NoteBox A

1, 2, 2, 3, 5

NoteBox B

2, 4, 4, 6

Find the probability for each of the following independent events. (Note: Probabilities are rounded to two decimal places where necessary).

Practice Questions

Part A: Both numbers are even

Find the probability that both drawn tickets have even numbers.

Correct Answer: 0.40

  1. Count the total outcomes and even numbers in each box:
    • Box A has 5 total numbers: 1, 2, 2, 3, 5. The even numbers are two twos. \(P(\text{Even from A}) = \frac{2}{5} = 0.40\)
    • Box B has 4 total numbers: 2, 4, 4, 6. All 4 numbers are even. \(P(\text{Even from B}) = \frac{4}{4} = 1.00\)
  2. Since the draws are independent, multiply their probabilities: \(P(\text{Both Even}) = P(\text{Even from A}) \times P(\text{Even from B}) = 0.40 \times 1.00 = 0.40\)

Part B: The sum of the numbers is exactly 6

Find the probability that the sum of the two drawn numbers equals 6.

Correct Answer: 0.25

  1. To get a sum of 6 we can either have (2,4) or (1,6).
  2. The chance to get (2,4) is \((2/5)*(2/4)=4/20\) and the chance to get (1,6) is \((1/5)*(1/4)=1/20\) by independence of draws from each box.
  3. Since these events are mutually exclusive the probabilities of getting (2,4) or (1,6) is $4/20+ 1/20 = 0.25.

Part C: The number from Box A is strictly greater than Box B

Find the probability that the number drawn from Box A is larger than the number drawn from Box B.

Correct Answer: 0.20

  1. Check each item in Box A against items in Box B to see where (A > B):
    • Box A elements: 1, 2, 2, 3, 5
    • Box B elements: 2, 4, 4, 6
  2. Out of 20 total combinations, only the pairs (3,2),(5,2),(5,4),(5,4) satisfy the requirement.
  3. Calculate probability: \(P(A > B) = \frac{4}{20} = 0.20\)

Part D: At least one of the numbers is a 2

Find the probability that at least one of the drawn tickets shows the number 2.

Correct Answer: 0.55

Using the complement rule is the easiest way to solve “at least one” problems: \(P(\text{At least one 2}) = 1 - P(\text{No 2s drawn})\)

  1. Find the probability of NOT drawing a 2 from each box:
    • Box A has 5 numbers, and 3 are NOT a 2 (1, 3, 5): \(P(\text{No 2 from A}) = \frac{3}{5}\)
    • Box B has 4 numbers, and 3 are NOT a 2 (4, 4, 6): \(P(\text{No 2 from B}) = \frac{3}{4}\)
  2. Multiply the probabilities to find the chance that neither box rolls a 2: \(P(\text{No 2s drawn}) = \frac{3}{5} \times \frac{3}{4} = \frac{9}{20} = 0.45\)
  3. Subtract from 1 to find the remaining probability: \(P(\text{At least one 2}) = 1 - 0.45 = 0.55\)