Data & Graphs

Problem

Probability is favorable over total outcomes

A fair die shows the numbers 1 through 6. For each of three described events, I find its probability as the fraction of favorable faces out of 6, then decide which event is the least likely.
Data & probability
Your answer
How to solve
Strategy Make a Systematic List — The sample space is just six small numbers, so the surest path is to list the faces that satisfy each event, count them, and write each probability as a fraction over 6. Once the three fractions are written with the same denominator, I simply check which one is smallest.
1STEP 1

List the whole sample space

The die can land on 1 through 6, all equally likely, so every chance is a count out of 6.

{1,2,3,4,5,6}→ 6 outcomes
2STEP 2

Count and write ㉠ (even)

The even faces are 2, 4, and 6 — three of them. So the probability of an even number is 3 out of 6.

P(even)=3/6=1/2
3STEP 3

Count and write ㉡ (multiple of 3)

The faces that are multiples of 3 are 3 and 6 — two of them. So this probability is 2 out of 6.

P(mult. of 3)=2/6=1/3
4STEP 4

Count and write ㉢ (6 or greater)

The only face that is 6 or greater is 6 itself — just one face. So this probability is 1 out of 6.

P( ≥ 6)=1/6
5STEP 5

Compare the three fractions

Over the same denominator 6 the numerators are 3, 2 and 1, so the last event is least likely.

3/6 > 2/6 > 1/6→ ㉢ lowest
Answer
Each probability is between 0 and 1 (3/6, 2/6, 1/6), which is exactly the range a probability must lie in, and the counts 3 + 2 ... use overlapping faces so they need not sum to 1. The fewest favorable faces (just 6) gives the smallest fraction, so ㉢ being lowest is consistent.
Takeaway

Count how many faces win, put it over 6, and the event with the fewest winning faces is the least likely.

  • List the whole sample space
  • Count and write ㉠ (even)
  • Count and write ㉡ (multiple of 3)
  • Count and write ㉢ (6 or greater)
  • Compare the three fractions
Where next?
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