Question

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 1

Answer:

Answer:Answer is 9.2

Answer:Answer is 9.2Step-by-step explanation:

Answer:Answer is 9.2Step-by-step explanation:Probability that the bomb will hit the target is 0.8

Answer:Answer is 9.2Step-by-step explanation:Probability that the bomb will hit the target is 0.8 n(s)= 0.8

Answer:Answer is 9.2Step-by-step explanation:Probability that the bomb will hit the target is 0.8 n(s)= 0.8Find the probability that out of ten bombs

Answer:Answer is 9.2Step-by-step explanation:Probability that the bomb will hit the target is 0.8 n(s)= 0.8Find the probability that out of ten bombs n(e)= 10

Answer:Answer is 9.2Step-by-step explanation:Probability that the bomb will hit the target is 0.8 n(s)= 0.8Find the probability that out of ten bombs n(e)= 10 P(e) = n(e)/n(s)

Answer:Answer is 9.2Step-by-step explanation:Probability that the bomb will hit the target is 0.8 n(s)= 0.8Find the probability that out of ten bombs n(e)= 10 P(e) = n(e)/n(s) 10 - 0.8

Answer:Answer is 9.2Step-by-step explanation:Probability that the bomb will hit the target is 0.8 n(s)= 0.8Find the probability that out of ten bombs n(e)= 10 P(e) = n(e)/n(s) 10 - 0.8 Answer is 9.2

Answer 2

$\large\underline{\sf{Solution-}}$

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.