The soldier fired until he hit the target. If the probability of hitting the target is 0.3, write the law of distribution of the number of shots.

Question

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Answer 1

There are two possibilities: the soldier hits the target (probability p=0.3); or he misses (probability q=0.7). There are only two possibilities so the distribution is binomial.

For a finite number of shots n, (p+q)ⁿ=(pⁿ+np^n-1q+n(n-1)p^n-2q²/2!+...(n!/(r!(n-r)!))p^n-rq^r+...n(n-1)p²q^n-2/2!+npq^n-1+qⁿ

When n=10, the terms on the right mean probability of all hits+9 hits out of 10 shots+8 hits out of 10 shots+...+2 hits out of 10 shots+1 hit out of 10 shots+no hits.

Probability has two parameters: n and r where r is the exact number of hits.

So p(n,r)=(n!/(r!(n-r)!))0.3^r0.7^n-r where 0≤r≤n. 

For example, if n=10 and r=3 (3 hits out of 10 shots)=p(10,3)=(10!/(3!7!))0.3³0.7⁷=120(0.027)(0.08235)=0.2668=26.68% approx.