The number of ways of arranging `6` players to throw the cricket ball so that oldest player may not throw first is

`P_(1) P_(2) P_(3) P_(4) P_(5) P_(6) P_(7) P_(8)` Given that if `P_(i), P_(j)` play with `i lt j`, then `P_(i)` will win. For the first round, `P_(4)` should be paired...

Given `S_(2)` reaches the semi-finals. Since all other players `(2^(n) - 1)` are equally likely to win the finals with probability p. `therefore (2^(n)-1)p+(1)/(4) = 1` `(2^(n) - 1)p =...

Correct Answer - D `H_("max")=(u^(2))/(2g)=40 rArr (u^(2))/(g)=80` `R_("max")=(u^(2))/(g)=80m`

Let A be the event of `P_(1)` winning in third round and B be the event of `P_(2)` winning in first round but ossing in second round. We have `P(A)(""^(8n-1)C_(n-1))/(""^(8n)C_(n))=1/8`...

Correct Answer - `a:b=2:1` Let W: evetn of drawing a white ball at any draw, and B: evetn of drawing a black ball at any draw Then `P(W)=(a)/(a+b)and P(B)=(b)/(a+b)` `P(A "wins...

Correct Answer - A For ranked 1 and 2 players to be winners and runners up respectively, they should not be paired with each other in any round. Therefore, the required...

Correct Answer - 3 For ranked 1 and 2 players to be winners and runners up, respectively, they should not be paired with each other in any round. `impliesp=30/31xx14/15xx6/7xx2/3=16/31`

Correct Answer - A `(a)` Since one player emerges the winner, each of the remaining `115` players loses in some round. So `115` matches are played. Alternatively , `58+29+14+7+4+2+1=115`

Correct Answer - A `(a)` First arrange `12` persons `A_(4),A_(5),….A_(15)` in `"^(15)P_(12)` ways There remains `3` places. Keep `A_(1)` in the first place and arrange `A_(2)`, `A_(3)` in the remaining two...

Correct Answer - A `(a)` Required number of ways `=("Number of ways in which" 16 "players can be divided in" `8` "couples")-("Number of ways when" S_(1) "and" S_(2) "are in the...