Question

Brandon is an amateur marksman. When he takes aim at a particular target on the shooting range, there is a 0.1, point, probability that he will hit it. One day, Brandon decides to attempt to hit 10 such targets in a row.

Assuming that Brandon is equally likely to hit each of the 10 targets, what is the probability that he will hit at least one of them?

Answer 1

Answer:

.65

Step-by-step explanation:

Strategy:

In this situation it is much easier to calculate the probability of the event we are looking for (he hits at least one target) by calculating the probability of its complement (he misses every target), and subtracting from 1.

In other words, we can use this strategy:

P(at least one hit)=1-P(miss all 10)

Calculations:

P(at least one hit)

=1-P(miss all 10)

=1-(0.9)^10

≈1-0.349

≈0.65

Answer:

P(at least one hit)≈0.65

I hope this helps!!!

Answer 2

The probability that Brandon will hit at least one of them is 0.65 approx.

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

$X \sim B(n,p)$

The probability that out of n trials, there'd be x successes is given by

$P(X =x) = \: ^nC_xp^x(1-p)^{n-x}$

For the considered case, as each hit is independent of each other, and there is either hit or not hit, so this situation can be modeled by binomial distribution.

For this case, we get:

• n = 10
• p = probability of success(we call it success if hit occurs) = 0.1
• X = count of successes in those 10 attempts

Then, we get: $X \sim B(n=10,p=0.1)$

The  probability that Brandon will hit at least one of them is written symbolically as:

$P(X \geq 1)$

We can rewrite it as:

$P(X \geq 1) = 1 - P(X < 1) = 1 - P(X = 0)$

Using the probability function of binomial distribution, we get:

$P(X \geq 1) = 1 - \: ^{10}C_0(0.1)^0(1-0.1)^{10} = 1 -0.9^{10} \approx 0.65$

Thus, the probability that Brandon will hit at least one of them is 0.65 approx.

Learn more about binomial distribution here:

https://brainly.com/question/13609688