VITEEE 2015 Paper

The probability of getting a double six in one throw of two dice = $\frac{1}{6} \times \frac{1}{6}$ = $\frac{1}{36}$ ∴ p = $\frac{1}{36}$ , q = 1 - p = 1 - $\frac{1}{36}$ = $\frac{35}{36}$ Now, $(p+q)^m$ = $q^n + \,{}^{n}C_{1} q^{n-1} p$ + $\,{}^{n}C_{2} q^{n-2} p^2$ + ... + $\,{}^{n}C_{r} q^{n-r} p^r$ + ... + $p^n$ The probability of getting atleast one double six in n throws with two dice. = $(q+p)^n - q^n$ = 1 - $q^n$ = 1 - $\left(\frac{35}{36}\right)^n$ ∴ 1 - $\left(\frac{35}{36}\right)^n$ > 0.99 ⇒ $\left(\frac{35}{36}\right)^n$ < 0.01 ⇒ n(log 35 - log 36) < log 0.01 ⇒ n[15441 - 15563] < -2 ⇒ - 0.0122n < - 2 ⇒ 0.0122n > 2 → n > 163.9 So, the least value of n is 164.