To solve the problem of ordering the given probabilities of Idris winning three different games from least likely to most likely, let's follow these steps:
1. Understand the Probabilities Given:
- The first probability is [tex]\( 47\% \)[/tex]. Percentages are easily converted to their decimal form for easier comparison. So, [tex]\( 47\% \)[/tex] is equivalent to [tex]\( 0.47 \)[/tex].
- The second probability is already given in decimal form: [tex]\( 0.52 \)[/tex].
- The third probability is given as a fraction: [tex]\( \frac{5}{10} \)[/tex]. Simplifying this fraction we get [tex]\( 0.5 \)[/tex].
2. List the Probabilities in Decimal Form:
- [tex]\( 0.47 \)[/tex] (from [tex]\( 47\% \)[/tex])
- [tex]\( 0.52 \)[/tex] (already in decimal form)
- [tex]\( 0.5 \)[/tex] (from [tex]\( \frac{5}{10} \)[/tex])
3. Order the Probabilities:
- To determine the order from least likely to most likely, we need to compare the decimal values.
- Comparing [tex]\( 0.47 \)[/tex], [tex]\( 0.5 \)[/tex], and [tex]\( 0.52 \)[/tex]:
- [tex]\( 0.47 \)[/tex] is less than [tex]\( 0.5 \)[/tex],
- [tex]\( 0.5 \)[/tex] is less than [tex]\( 0.52 \)[/tex].
4. Final Order from Least Likely to Most Likely:
- Least likely: [tex]\( 0.47 \)[/tex]
- Next: [tex]\( 0.5 \)[/tex]
- Most likely: [tex]\( 0.52 \)[/tex]
Therefore, the probabilities of Idris winning the games, ordered from least likely to most likely, are:
[tex]\[ 0.47, 0.5, 0.52 \][/tex]
