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**Hint:**The probability of hitting a target in any shot (p) is given. Since the target is one, the probability of not hitting the target in any shot (q) will be $1 - {\text{p}}$. The probability of hitting a target twice will be( $1 - {\text{target hit once - target hit never}}$) .

**Complete step-by-step answer:**Given that the probability of hitting a target in any shot (p) =$0.2$

Fired shots =$10$ so there are $10$ chances of hitting the target. We have to find the probability that the target will be hit at least twice.

Since the target is one so the probability of not hitting the target in any shot (q) =$1 - {\text{p}}$[as p + q=$1$ ]

So q=$1 - 0.2 = 0.8$

Now, if $10$ shots are fired and the target is hit once means that the target was not hit the $9$ times.

So the probability of target hit once=the probability of hitting target one time× (the probability of not hitting a target $9$ times) =${{\text{p}}^1}.{{\text{q}}^9} = 0.2 \times {\left( {0.8} \right)^9}$ --- (i)

And the probability of target never hit =${{\text{q}}^{10}} = {0.8^{10}}$ --- (ii)

The probability of the target hit twice will be = $1 - {\text{target hit once - target hit never}}$

On putting the values from eq. (i) and (ii) in the formula we get,

$ \Rightarrow $ The probability of the target hit twice=$1 - \left[ {0.2 \times {{0.8}^9}} \right] - {0.8^{10}}$

On taking ${0.8^9}$ common we get,

$ \Rightarrow $ The probability of the target hit twice=$1 - \left( {{{0.8}^9}} \right)\left( {0.8 + 0.2} \right) = 1 - {0.8^9}$

On solving further we get,

$ \Rightarrow $ The probability of the target hit twice=$1 - 0.134217 = 0.865783$

**Hence the answer is $0.865783$.**

**Note:**Here the student may get confused in eq. (i). Since the probability of not hitting a target in any shot is given, so if the target hit once in $10$ shots then in $9$ shots it will miss. And the probability of not hitting a target in a shot is $0.8$ then the probability of not hitting the target in $9$ shots will become ${0.8^9}$ .

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