Statement -1 : The probability of getting a tail most of the times in 10 tosses of a unbiased coin is `(1)/(2){1-(10ǃ)/(2^10 5ǃ5ǃ)}`.
Statement -2: `.^2nC_(0)+.^2nC_(1)+.^2nC_(2)+.....^2nC_(n)=2^2n-1,n in N`.

Options

A. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

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)=.^10C_(r)((1)/(2))^10`
`therefore` Probability of getting a tail most of the times `=P(Xge 6)=sum_(r=6)^(10)P(X=r)=(1)/(2)^10{.^C_(6)+.^C_(7)+.^10C_(8)+.^10C_(9)+.^10C_(10)}`
`=((1)/(2))^11{.^10C_(0)+.^10C_(1)+.... .^10C_(10)-.^10C_(5)}`
`=((1)/(2))^11{2^10-(10ǃ)/(5ǃ 5ǃ)}=(1)/(2){1-(10ǃ)/(2^10 5ǃ5ǃ)}`
So, statement -1 is true.
We have,
`.^2nC_(0)+.^2nC_(1)+.^2nC_(2)+.....+.^2nC_(n)+.^2nC_(n+1)+.......+.^2nC_(2n)=2^2n`
`rArr 2(.^C_0+.^2nC_1+.^2nC_2+..... .^2nC_(2)+....+.^2nC_(n-1))+.^2nC_(n)=2^2n`
`rArr .^2nC_0+.^2n_C_1+.^2nC_(2)+......^2nC_n-1=2^2n-1-(1)/(2)2n_(C_n)`
So, Statement -2 is not true.