# The probability that a marksman will hit a target is 4/5. He fires 10 shots. Calculate, correct to 3 decimal places, the probability that he will hit the target if he hits the target exactly 7 times, calculate the probability that the 3 misses are with successive shots.

You need to use binomial random variable which has probability density function of` `

`f(k;n,p)=P(X=k)=((n),(k))p^k(1-p)^(n-k)`

Let's now explain what above formula means. It says that probability of getting `k` successes out of `n` Bernoulli trials is equal to product of `p^k` (`k` successes), `(1-p)^(n-k)` (`n-k` fails) and `((n,),(k,))` (in how many ways can we choose combinations of successes).

In the...

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You need to use binomial random variable which has probability density function of` `

`f(k;n,p)=P(X=k)=((n),(k))p^k(1-p)^(n-k)`

Let's now explain what above formula means. It says that probability of getting `k` successes out of `n` Bernoulli trials is equal to product of `p^k` (`k` successes), `(1-p)^(n-k)` (`n-k` fails) and `((n,),(k,))` (in how many ways can we choose combinations of successes).

In the first case we use probability of opposite event i.e. probability that a marksman will hit the target at least one time is opposite event of not hitting target even once:

`P(X geq 1)=1-P(X=0)=1-((10),(0))(4/5)^0(1/5)^10`

`P(X geq 1)=1-0.0000001024=0.9999998976`  <-- Solution

In second case we know that the target has been hit 7 times so we don't need to worry about that. The question is in how many ways can we arrange 7 hits and 3 misses so that the 3 misses are successive. That is the same as asking in how many ways can we choose block of 3 successive elements out of 10. We can do that in

`((n-(k-1)),(1))=((10-(3-1)),(1))=8`

ways. Now we only need to divide that number by the total number of arrangements of 7 hits and 3 misses which is

`(n!)/(k_1!k_2!)=(10!)/(7!3!)=120`