Question

An unbiased die with faces marked 123456 thrown n times and the list n numbers showing up is noted what is the probability thatamong numbers 123456 only three show up in the list

Answer 1

Please find the answer to your questionThe total no. of outcomes = 6nWe can choose three numbers out of 6 in 6 C3 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= 3 C2 [2n – 1n – 1n] = 3(2n – 2) and the number of n sequence which are exactly one number= (3 C1) (In) = 3Thus, the number of sequences, which use exactly three numbers= 6 C3 [3n – 3(2n – 2) – 3] = 6 C3 [3n – 3(2n) + 3]∴ Probability of the required event,= 6 C3 [3n – 3(2n) + 3]/6n
this i the first way u can do
or u can do it this way
 

 Favourable points are 2,3,4,5∴P=46=23∴P=46=23Since the die is rolles 4 times∴∴ Required probability = (23)4(23)4=1681


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