The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least $0.96$, is

$\text{Given: fair coin, } P(H) = P(T) = \dfrac{1}{2}$

$n = \text{number of tosses}, \quad P(\text{at least 2 heads}) \geq 0.96$

$\text{To find: minimum } n \text{ with } P(X \geq 2) \geq 0.96$

$P(X \geq 2) = 1 - P(X=0) - P(X=1)$ $\text{Require } P(X \geq 2) \geq 0.96$ $P(X=0) = \left(\dfrac{1}{2}\right)^{n}$ $P(X=1) = n\left(\dfrac{1}{2}\right)^{n}$ $P(X=0) + P(X=1) = \dfrac{n+1}{2^{n}}$ $P(X \geq 2) = 1 - \dfrac{n+1}{2^{n}}$ $1 - \dfrac{n+1}{2^{n}} \geq 0.96 \iff \dfrac{n+1}{2^{n}} \leq 0.04$ $\text{Need } \dfrac{n+1}{2^{n}} \leq 0.04$ $n=7:\ \dfrac{8}{128} = \dfrac{1}{16} = 0.0625 > 0.04\ \text{✗}$ $n=8:\ \dfrac{9}{256} = 0.0352 < 0.04\ \checkmark$ $P(X \geq 2) = 1 - \dfrac{9}{256} = \dfrac{247}{256} \approx 0.9648 \geq 0.96\ \checkmark$ $\text{Minimum number of tosses} = 8$