`Aa n dB`
play a
series of games which cannot be drawn and `p , q`
are their respective chance of winning a single game. What
is the chance that `A`
wins `m`
games
before `B`
wins `n`
games?
For this to happen, A must win at least m out of the first m + n - 1 games. Therefore, the required probability is `""^(m+n-1)C_(m)p^(m)q^(n-1)+""^(m+n-1)C_(m+1)p^(m+1)q^(n-2)+...+""^(m+n-1)C_(m+n-1)p^(m+n-1)`
Correct Answer - B
Total number of ways of distribution is `4^(5)`.
`therefore n(S) = 4^(5)`
Total number of ways of distribution so that each child gets at least one game...
Correct Answer - B
Team totals must be 0, 1, 2, …., 39. Let the teams be `T_(1)T_(2),…, T_(40)`, so that `T_(i)` loses to `T_(j)` for `i lt j`. In other...
Correct Answer - A::C
Let `p_(1), p_(2)` be the chances of happening of the first and second events, respectively. Then according to the given conditions, we have
`p_(1) = p_(2)^(2)`
and...
`P_(1)` can win in the following matually exclusive ways:
(a) `P_(1)` wins the next dix matches.
(b) `P_(1)` wins five out of next six matches, so that after new six...
Correct Answer - Fisth offer
The probability `p_(1)` of winning the best of three games is equal to the sum of the probability of swimming two games and the probability of...
Correct Answer - A
`P(a)=0.3P(b)=0.5,P(c)=0.2.` Hence, a,b c are exhaustive. P(same horse wins all the three reces)
=P(aaa or bbb or ccc)
`=(0.3)^(3)+(0.5)^(3)+(0.2)^(3)`
`=(27=125+8)/(1000)=(160)/(1000)=4/25`
P(each horse wins exactly one race)
=P...
Correct Answer - B
`P(XgtY)=T_(1)T_(1)+DT_(1)+T_(1)D("where"T_(1)" represents wins and D represents draw")`
`=((1)/(2)xx(1)/(2))+((1)/(6)xx(1)/(2))+((1)/(2)xx(1)/(6))=5/12`
Correct Answer - C
`(c )` Player `A` can win if `A` throws `(1,6)` or `(6,1)` and `B` throws `((1,1),(2,2),(3,3),(4,4),(5,5) or (6,6))`. Thus the number of ways is `12`.
Similarly the...