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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=$1 - 0.134217 = 0.865783$