Question

The probability that man can hitting of target is 1/4 if he fires 5 times, whais in a probability of his hitting target exactly 3 times​

Answer 1

Step-by-step explanation:

The probability of something to happen is independent of the number of experiments you conduct. So can we say that the shooter hits the target with p=.25 - this is probably what you meant. Now, he shoots 7 times.

Mathematically speaking, we have a Bernoulli-process of length 7 with p = 0.5. Let random variable X be the number of hits he got from the 7 shots.

We want to know

r = B( 0.25; 7 )( X>=2 ) ; r is Probability to hit at least 2

B( 0.25; k)( X >=1 ) = 2/3 ; k is number of shots needed to have at least 1 hit wit probability 2/3

Type -1 - calculations can be done with one of the many calculators on the internet. I used this one here: 1 picture

So the probability to hit at least 2 is about 55%

So the probability to hit at least 2 is about 55%For Type - 2- questions, I did not find a calculator in the internet. We have to solve the equation manually which I will not do here because for some reason I cannot write mathematical formula in Quora.

So the probability to hit at least 2 is about 55%For Type - 2- questions, I did not find a calculator in the internet. We have to solve the equation manually which I will not do here because for some reason I cannot write mathematical formula in Quora.But we can do a little try and error, and manipulate the number of trials until we get near 0.66 as resulting probability for X >= 1.

So the probability to hit at least 2 is about 55%For Type - 2- questions, I did not find a calculator in the internet. We have to solve the equation manually which I will not do here because for some reason I cannot write mathematical formula in Quora.But we can do a little try and error, and manipulate the number of trials until we get near 0.66 as resulting probability for X >= 1.This comes closest: 2 pic

He must fire 4 times to get at least one hit with a probability of 68,3%, (which is the nearest value to 2/3)

Answer 2

Answer:

The probability of $hitting$ the target exactly $3$ times is $\frac{45}{512}$.

Step-by-step explanation:

Formula for binomial distribution probability:

$P(X=x)=n_C_xa^xb^{n-x}$, where

$P$ = binomial probability,

$x$ = number of times for an outcome within $n$ trails,

$n_C_x$ = number of combinations

$a$ = Probability of $success$ on a single trial

$b$ = Probability of $failure$ on a single trial

$n$ = number of trials

According to the question,

The probability of a man $hitting$ a target = $\frac{1}{4}$

i.e., probability of $success$, $a$ = $\frac{1}{4}$

So, the probability of not $hitting$ the target, $b$ = $1-\frac{1}{4}$

$b=\frac{3}{4}$

He fires $9$ times, i.e.,

Number of trials, $n$ = $5$

The probability of $hitting$ the target exactly $3$ times = $P(X=3)$

= $5_C_3\left(\frac{1}{4}\right) ^3\left(\frac{3}{4}\right)^{5-3}$

= $\frac{5!}{3!(5-3)!} \left(\frac{1}{4}\right) ^3\left(\frac{3}{4}\right)^{2}$

= $\frac{5\times 4\times 3!}{3!2!}\cdot \frac{3^2}{4^5}$

On simplifying, we get

= $\frac{90}{1024}$

= $\frac{45}{512}$

Thus, the probability of $hitting$ the target exactly $3$ times is $\frac{45}{512}$.

#SPJ3