probability that the bomb will hit the target is 0.8 find the probabil

1.	probability that the bomb will hit the target is 0.8 find the probability that out of ten bombs dropped ,excetly two will miss the target
Answer» $\large\underline{\sf{Solution-}}$ We know, By Binomial Distribution, $\boxed{ \bf \: P(r) = \:^n C_r {p}^{r} {q}^{n - r} }$ where, n = number of trials p = probability of SUCCESS q = probability of failure r = random variable whom success is desired. and p + q = 1. Let's SOLVE the problem now!! Here, given that Number of bombs dropped, n = 10. Probability that BOMB hit the target, p = 0.8 So, Probability that bimb will not hit target, q = 1 - 0.8 = 0.2 Now, We have to find the probability exactly 2 bombs will miss the target. $\bf :\longmapsto\:P(exactly \: 2 \: bombs \: miss \: the \: target)$ $\: \: \bf \: = \: P(exactly \: 8 \: bombs \: hit \: the \: target)$ $\: \: = \sf \: \:^{10} C_8 \: \times {(0.8)}^{8} \times {(0.2)}^{2}$ $\: \: = \sf \: \dfrac{10 \times 9}{2 \times 1} \times {\bigg( \dfrac{8}{10} \bigg) }^{8} \times {\bigg( \dfrac{2}{10} \bigg) }^{2}$ $\: \: = \sf \: 45 \times {\bigg(\dfrac{4}{5} \bigg) }^{8} \times {\bigg( \dfrac{1}{5} \bigg) }^{2}$ $\: \: = \sf \: 45 \times \dfrac{ {4}^{8} }{ {5}^{8} } \times \dfrac{1}{ {5}^{2} }$ $\: \: = \sf \: \dfrac{9 \times {2}^{16} }{ {5}^{9} }$ Additional Information :- If the following conditions are satisfied, then X has a binomial distribution with parameters n and p, represented as B(n,p). 1. The number of observations n is fixed. 2. Each observation is independent. 3. Each observation represents ONE of two outcomes (succes, p or failure, q). 4. The probability of "success" p is the same for each OUTCOME.