An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown n tim

1.	An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown n times and the list of n numbers showing up is noted. What is the probability that, among the numbers 1, 2, 3, 4, 5, 6, only three numbers appear in this list?
Answer» The total no. of outcomes = 6ⁿ We can choose three numbers out of 6 in ⁶C₃ ways. By using three numbers out of 6 we can get 3ⁿsequences of length n. But these sequences of length n which use exactly two numbers and exactly one number. The number of n – sequences which use exactly two numbers = ³C₂ [2ⁿ– 1ⁿ – 1ⁿ] = 3(2ⁿ – 2) and the number of n sequence which are exactly one number = (³C₁) (In) = 3 Thus, the number of sequences, which use exactly three numbers = ⁶C₃ [3ⁿ – 3(2ⁿ – 2) – 3] = ⁶C₃ [3ⁿ – 3(2ⁿ) + 3] ∴ Probability of the required event, = ⁶C₃[3ⁿ – 3(2ⁿ) + 3]/6ⁿ

An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown n times and the list of n numbers showing up is noted. What is the probability that, among the numbers 1, 2, 3, 4, 5, 6, only three numbers appear in this list?

Answer»

The total no. of outcomes = 6ⁿ

We can choose three numbers out of 6 in ⁶C₃ ways. By using three numbers out of 6 we can get 3ⁿsequences of length n. But these sequences of length n which use exactly two numbers and exactly one number.

The number of n – sequences which use exactly two numbers

= ³C₂ [2ⁿ– 1ⁿ – 1ⁿ] = 3(2ⁿ – 2) and the number of n sequence which are exactly one number

= (³C₁) (In) = 3

Thus, the number of sequences, which use exactly three numbers

= ⁶C₃ [3ⁿ – 3(2ⁿ – 2) – 3] = ⁶C₃ [3ⁿ – 3(2ⁿ) + 3]

∴ Probability of the required event,

= ⁶C₃[3ⁿ – 3(2ⁿ) + 3]/6ⁿ

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