Probability

Question

any one can solve? 

Comments

prob. of success = 4/7 

prob. of failure = 1-4/7 =3/7

since they have specified in question that he succeed twice out of 10

So first of select 2 shot out of 10 shots, number of ways = C(10,2) 

According to Binomial distribution =  ¹⁰c₂ (4/7)² (3/7)⁸

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How power 2 and 8

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exactly twice he hits the target that's why 2 and remaining 8 times he doesn't hit the target..

Answer 1

in this question there is probability of success given (shot on target ) p=$\frac{4}{7}$ 

and probability of failure(miss the shot) is q=1-$\frac{4}{7}$ 

                                                                    =$\frac{3}{7}$ 

so clearly question is asking out of 10 shot 2 shot on the target what is the probability of that which is binomial distribution.

the formula for the binomial distribution is $\binom{n}{r}$ = ⁿC_r*(p)^(r)*q^(n-r)

so putting in the formula we have   ¹⁰C₂*(4/7)²*(3/7)^{8  which is the answer.}

Comments

isn't it possible to solve using probability theory directly ?

Answer 2

 We know that, Binomial distribution  = ⁿC_r*(p)^(r)*q^(n-r)

According to question ,     probability of success (p) = 4/7 ,        probability of failure(q) = 1 - 4/7 = 3/7 

Given,    n=10, r=10 

Probabilty that person hit the target twice = ¹⁰c₂ (4/7)² (3/7)⁸              ans