What is the minimum number of times a fair coin must be tossed so that the probability of getting at least one head exceeds 0.95?

If the coin is tossed once, there will be $2^1 = 2$ choices i.e. $=\{H, \boxed{T}\}$

If the coin is tossed twice, there will be $2^2 = 4$ choices i.e. $=\{HH, HT, TH, \boxed{TT}\}$

If the coin is tossed thrice, there will be $2^3 = 8$ choices i.e. $=\{HHH, HHT, HTH, HTT, THH, THT, TTH, \boxed{TTT}\}$

There will only be one case of "all tails" (as shown in the boxed outcomes above).

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Using the complement:

$P(\text{at least one head}) = 1 - P(\text{all tails})$

The probability of "at least one head" must exceed $0.95$, so:

$1 - P(\text{all tails}) > 0.95$

$P(\text{all tails}) < 0.05$

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Let $n$ = number of coin tosses

$n = 3$: Probability of all tails $=\dfrac{1}{8} = 0.125$

$n = 4$: Probability of all tails $=\dfrac{1}{16} = 0.0625$

$n = 5$: Probability of all tails $=\dfrac{1}{32} \approx 0.03125$

Since $0.03125 < 0.05$, the condition is satisfied.

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Therefore, the minimum number of tosses is $5$.