# Toss a fair die four times. What is the probability that all tosses produce different outcomes?

Toss a fair die four times. What is the probability that all tosses produce different outcomes?

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Number of possible outcomes
Fundamental counting principle: If the first event could occur in m ways and the second event could occur in m ways, then the number of ways that the two events could occur in sequence is m - 1.
A fair die has 6 possible outcomes: 1, 2, 3, 4, 5, 6. The fair die is rolled 4 times.
First roll: 6 ways
Second roll: 6 ways
Third roll: 6 ways
Fourth roll: 6 ways
Use the fundamental counting principle: # of possible outcomes = $$\displaystyle{6}\cdot{6}\cdot{6}\cdot{6}={6}^{{4}}={1296}$$
Number of favorable outcomes
Fundamental counting principle: If the first event could occur in m ways and the second event could occur in m ways, then the number of ways that the two events could occur in sequence is m - 1.
A fair die has 6 possible outcomes: 1, 2, 3, 4, 5, 6. The fair die is rolled 4 times, We require the outcomes of each roll to be different
First roll: 6 ways
Second roll: 5 ways
Third roll: 4 ways
Fourth roll: 3 ways
Use the fundamental counting principle:
# of possible outcomes=$$\displaystyle{6}\cdot{5}\cdot{4}\cdot{3}={360}$$
Probability
The probability is the number of favorable outcomes divided by the number of possible outcomes:
P(Different outcomes)=# of favorable outcomes/# of possible outcomes $$=\frac{360}{1296} =\frac{5}{18}\sim0.2778=27.78\%$$