Question

A symmetric die is thrown $(2n+1)$ times. The probability of getting a prime score on the upturned face at most $n$ times is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{2}{3}$

Answer 1

Answer: (A)

$\frac{1}{2}$

Probability of getting prime no. in single throw $=1/2$
Probability of not getting prime no. in single throw $=1/2$
Probability of getting $n$ throws ${=}^{2n+1}{C}_0\ast \left({1/2}^0\right)\ast {(1/2}^{2n+1}){+}^{2n+1}{C}_1\ast ({1/2}^1)\ast {(1/2}^{2n})+.......{.}^{2n+1}{C}_{n-1}\ast \left({1/2}^{n-1}\right)\ast {(1/2}^{n+2}){}^{}{+}^{2n+1}{C}_n\ast ({1/2}^n)\ast {(1/2}^{n+1})={1/2}^{2n+1}{(}^{2n+1}{C}_0{+}^{2n+1}{C}_1+.......{.}^{2n+1}{C}_{n-1}{+}^{2n+1}{C}_n)={1/2}^{2n+1}\ast 2^{n+1}/2=1/2$
Sum of ${}^{2n+1}{C}_0{+}^{2n+1}{C}_1+.......{.}^{2n+1}{C}_{n-1}{+}^{2n+1}{C}_n$ can be found by putting $x=1$ in binomial expansion of $(1+x{)}^{(}2n+1)$ and is equal to half of that sum.
Hence, option 'A' is correct.