The number of `n` digit number formed by using digits `{1,2,3}` such that if `1` appears, it appears even number of times, is
A. `2^(n)+1`
B. `(1)/(2)(3^(n)+1)`
C. `(1)/(2)(3^(n)-1)`
D. `(1)/(2)(2^(n)-1)`
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 - 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 - `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
According to the givn condition,
`""^(n)C_(3)((1)/(2))^(n)=""^(n)C_(4)((1)/(2))^(n),`
where n is the number of times die is thrown.
`therefore""^(n)C_(3)=""^(n)C_(4)impliesn=7`
Thus, the required probability is
`=""^(7)C_(1)((1)/(2))^(7)=(7)/(2^(7))=(7)/(128)`
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)` Total number of ways such that at least `3` digits will not occur in its position.
`=^(5)C_(3){3!-^(3)C_(1)2!+^(3)C_(2)1!-^(3)C_(2)0!}+^(5)C_(4)[4!-^(4)C_(1)(3!)+^(4)C_(2)(2!)-^(4)C_(3)(1!)+^(4)C_(4)(0!)}+^(5)C_(5){5!-^(5)C_(1)4!+^(5)C_(2)3!-^(5)C_(3)2!+^(5)C_(4)1!-^(5)C_(5)(0!)}`
`=10(2)+5(9)+(44)`
`=20+45+44=109`
Correct Answer - C
`(c )` `E:` Event of getting on outcome `(n_(1), n_(2),n_(3))` such that `i^(n_(1))+i^(n_(2))+i^(n_(3))=1`
`E_(1) : ` Event that fair die is thrown
`E_(2) :` Event that biased...
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