A binary number is made up to 8 digits. Suppose
that the probability if an incorrect digit appearing is `p`
and that the errors in different digits are independent of each other. Then find the probability of
forming an incorrect number.
Correct Answer - `1-(1-p)^(8)`
Probability for an incorrec digit is p. hence, probability for 8 correct digit is `1-p)^(8).` Hence, required probability is `1-(1-p)^(8).`
Correct Answer - D
Three-digit numbers are 100, 101, …, 999. Total number of such numbers is 900. The three-digit numbers (which have all same digits) are 111, 222, 333, …,...
Correct Answer - B
`(b)` Least digit used `=4`
`:.` We can use `4,5,6,7,8,9`.
So each of the four places can be filled in `6` ways
So number of numbers is...
Correct Answer - C
`(c )` Case I : All six digits alike
i.e. `111111`, `222222`……..etc. `=5` ways
Case II : `2` alike `+2` other alike.
Select any three in `"^(5)C_(3)`...
Correct Answer - B
`(b)` Required number of numbers `="(n)C_(0)(2)^(n)+"(n)C_(2)(2)^(n-2)+…`
We know that
`(1+x)^(n)+(x-1)^(n)=2sum_(k=0)^(n)"(n)C_(k)x^(k)`
Hence, put `x=2`, get the result
`2["^(n)C_(0)(2)^(n)+^(n)C_(2)(2)^(n-2)+^(n)C_(2)(2)^(n-4)+^(n)C_(6)(2)^(n-6)+...]`
Correct Answer - A
`(a)` Number of digits are `9`
Select `2` places for the digit `1` and `2` in `"^(9)C_(2)` ways from the remaining `7` places, select any two places...