Doubt

this is an iit q.

Starting with N₀ nuclei, the number of nuclei at the end of time t is
N=N_0e^{-\lambda t}
where \lambda = \dfrac{\ln 2}{T_{1/2}}
So the number of nuclei disintegrated till time t equals N_0(1-e^{\lambda t})
So the probability that a nucleus disintegrates after time t is
p(t)=\dfrac{N_0(1-e^{\lambda t})}{N_0}=1-e^{-\lambda t}
For the present problem, t = 2 T_1/2
Thus,
p(2T_{1/2})=1-e^{-\lambda 2T_{1/2}}=1-e^{-2\ln 2}=1-\dfrac{1}{4}=\dfrac{3}{4}
So Nishant sir's answer is indeed correct.

Answer IS given 3/4![3]

i have fiitjee archive,,there its given 1/2