Correct Answer - C
Let p be the probability of getting a tial in a sigle trial. Then, p
1
2
=q. Let X denote the number of tails in 10 trials. Then,
-P(X=r)=.10Cr(
1
2
)10
∴ Probability of getting a tail most of the times =P(X≥6)=
10
∑
r=6
P(X=r)=
1
(2)10
{.C_(6)+.C_(7)+.10C8+.10C9+.10C10}
=(
1
2
)11{.10C0+.10C1+.....10C10-.10C5}
=(
1
2
)11{210-
10ǃ
5ǃ5ǃ
}=
1
2
{1-
10ǃ
2105ǃ5ǃ
}
So, statement -1 is true.
We have,
.2nC0+.2nC1+.2nC2+.....+.2nCn+.2nCn+1+.......+.2nC2n=22n
⇒2(.C_0+.2nC1+.2nC2+......2nC2+....+.2nCn-1)+.2nCn=22n
⇒.2nC0+.2nC_1+.2nC2+......2nCn-1=22n-1-
1
2
2nCn
So, Statement -2 is not true.
