Probability of hitting the “bullseye” [closed]

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I'm trying to solve this task but I'm obviously stuck. Will anyone help me solve it?

John is throwing darts at a target. He is very good at the game, so the chances of him hitting the “bullseye” (i.e. the center) are $25$%. It is reasonable to assume that individual attempts are independent. If he throws $5$ times in total, what is the probability that he will hit the bullseye:
a) Exactly three times
b) At least three times

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1 Answer

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This is a binomial distribution since these are independent Bernoulli trials so a trial outcome is either success (hitting the bullseye) or failure and you want the probability of getting a certain number of successes in a certain number of trials regardless of where they happen.

In Bionmial distribution $P(X=k)$ is of the form:

$\binom{n}{k}p^k(1-p)^{n-k}$

Where:

$p$ is the probability of success
$n$ is the total number of trials
$k$ is the number of successes

The general formula for your question would be:

$\binom{5}{k}\cdot (0.25)^k \cdot (0.75)^{5-k}$For your first question we get:$\binom{5}{3}\cdot (0.25)^3 \cdot (0.75)^{2}$

For your second question they are asking for the probability that AT LEAST $3$ successes so that is the probability of getting three successes+ the probability of getting four successes+ probability of getting four successes, we get the following sum:

$\sum_{k=3}^5 \binom{5}{k}\cdot (0.25)^k \cdot (0.75)^{5-k}$

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