Answer:
0.0384
Step-by-step explanation:
Given a fair 5 sided die rolled 5 times.
Numbers on it are 1, 2, 3, 4, 5.
To find:
The probability that each number will occur exactly once.
Solution:
Formula for probability of an event E:
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}[/tex]
At the first roll of die, any number can occur.
So number of possible outcomes = 5
Total number of possible outcomes = 5
[tex]P(1^{st}\ roll) = \dfrac{5}{5} = 1[/tex]
At the second roll of die, any number can occur other than that occurred in first roll.
So number of possible outcomes = 4
Total number of possible outcomes = 5
[tex]P(2^{nd}\ roll) = \dfrac{4}{5}[/tex]
At the third roll of die, any number can occur other than that occurred in first and second roll.
So number of possible outcomes = 3
Total number of possible outcomes = 5
[tex]P(3^{rd}\ roll) = \dfrac{3}{5}[/tex]
At the fourth roll of die, any number can occur other than that occurred in first, second and third roll.
So number of possible outcomes = 2
Total number of possible outcomes = 5
[tex]P(4^{th}\ roll) = \dfrac{2}{5}[/tex]
At the fifth roll of die, any number can occur other than that occurred in first, second, third and fourth roll.
So number of possible outcomes = 1
Total number of possible outcomes = 5
[tex]P(5^{th}\ roll) = \dfrac{1}{5}[/tex]
The required probability will be multiplication of all the five probabilities.
[tex]1 \times 0.8 \times 0.6 \times 0.4 \times 0.2 = \bold{0.0384}[/tex]
