An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6 is thrown 6 times and the list of the 6 numbers showing up is noted. If $P (A_{i})$ denotes the probability that only i number of numbers appear in the list out of numbers 1, 2, 3, 4, 5, 6 then which of the following is/are CORRECT

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$P (A_{3}) > P (A_{4})$

$P (A_{4}) = \frac{23400}{6^{6}}$

$\sum_{i = 1}^{6} P (A_{i}) < 1$

$\max (P (A_{i})) = P (A_{4})$

answer is B, D.

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Detailed Solution

$P (A_{3}) = \frac{^{6} C_{3} \times (\frac{6!}{2! 2! 2!} + \frac{6!}{1! 2! 3!} \times 3! + \frac{6!}{1! 1! 4!} \times 3)}{6^{6}} = \frac{10800}{6^{4}}$

$P (A_{4}) = \frac{^{6} C_{4} (\frac{6! \times 4!}{3! {(1!)}^{3} 3!} + \frac{6! 4!}{{(2!)}^{4} {(1!)}^{2}})}{6^{6}} = \frac{23400}{6^{6}}$

$P (A_{5}) = \frac{^{6} C_{5} \times (^{5} C_{1} \times \frac{6!}{2!})}{6^{6}} = \frac{10800}{6^{6}}$

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