One ticket is selected at ransom form 50 tickets numbered `00,01,02,…,49.` Then the probability that the sum of the digits on the selected ticket is 8

Correct Answer - `(.^(29)C_(2)xx.^(20)C_(2))/(.^(50)C_(5))` Five tickets out of 50 can be drawn in `.^(50)C_(5)` ways. Since `x_(1) lt x_(2) lt x_(3)lt x_(4) lt x_(5)` and `x_(3) = 30`, we have `x_(1),...

Correct Answer - Not possible P(A) = 0.5, `P(A nn B) le 0.3` So, `P(A uu B) = P(A) + P(B) - P(A nn B)` gives `P(B) = P(A uu B)...

Correct Answer - B The prime digits are 2, 3, 5, 7. If we fix 2 at first place, then other 2n - 1 places are filled by all four digits....

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 - A Let the number selected be xy. Then `x+y=9,0lex,yle9` `and xy=0impliesx=0,y=9or y=0,x=9` `P(x_(1)=9//x_(2)=0)(P(x_(1)=9nnx_(2)=0))/(P(x_(2)=0))` Now, `P(x_(2)=0)=19/100` `andP(x_(1)=9nnx_(2)=0)2/100` `impliesP(x_(1)=9//x_(2)=0)=(2//100)/(19//100)=2/19`

Correct Answer - B The sum of the digits can be 7 in the following ways: `07,16,25,34,43,52,61,70.` `therefore(A=7)={07,16,25,34,43,52,61,70}` Similarly, `(B=0)={00,01,02...,10,20,30,...,90}` Thus, `(A=7)nn(B=0)={09,70}` `thereforeP((A=7)nn(B=0))=2/100,P((B=0))=19/100` `P(A=7)|B=0=(P((A=7)nn(B=0)))/(P(B=0))` `=(2/100)/(19/100)=2/19`

Correct Answer - D Let X denote the largest number on the 3 sticks drawn. Then, `P(Xle7)=(7//20)^(3)and P(Xle6)=(6//20)^(3).` Then, the required probability is `P(X=7)=((7)/(20))^(3)-((6)/(20))^(3)`

Correct Answer - A `(a)` `1234ul(5678910)` `1^(st)` drawn is `5`, then `2^(nd)` drawn can be `1` only. If `1^(st)` is `6`, then `2^(nd)` is `1` or `2` and so on. ltbr...

Correct Answer - `(43)/(50)`

Correct Answer - f=300x+800y