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LekaFEV [45]
3 years ago
13

Find the probability of exactly four successes in five trials of a binomial experiment in which the probability of success is 40

%. Round to the nearest tenth of a percent.
Mathematics
2 answers:
hodyreva [135]3 years ago
8 0

Answer:

p(4 successes) ≈ 7.7%

Step-by-step explanation:

p(k successes in n trials) = C(n,k)·p^k·(1-p)^(n-k)

You have n=5, k=4, p=0.4, so the probability is ...

p(4 successes) = C(5,4)·(0.4)^4·(1 -0.4)^1 = 5·0.4^4·0.6 = 0.0768

p(4 successes) ≈ 7.7%

Igoryamba3 years ago
6 0

Answer:

P(4 success)=0.0768

Step-by-step explanation:

The Probability for r success in binomial distribution is given by:

P(r success)=n_{C_{r}}p^rq^{n-r}

Where n is the number of trails

p is the probability of success in each trial

q =1-p which is the probability of failure

Here r=4

n=5

p=40%=0.4 and q=1-0.4=0.6

Hence, P(4 success)= 5_{C_{4}}0.4^40.6^{5-4}

                                 =  5\times 0.4^4\times 0.6

                                 = 0.0768

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The city has an average of 13 days of rainfall for April.
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Using the Poisson distribution, we have that:

  • There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.
  • There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.
  • There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

<h3>What is the Poisson distribution?</h3>

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}

The parameters are:

  • x is the number of successes
  • e = 2.71828 is the Euler number
  • \mu is the mean in the given interval.

For this problem, the mean is given as follows:

\mu = 13

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:

P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}

P(X = 10) = \frac{e^{-13}13^{10}}{(10)!} = 0.0859

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

The probability of having less than three days of precipitation in the month of April is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which:

P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-13}13^{0}}{(0)!} \ approx 0

P(X = 1) = \frac{e^{-13}13^{1}}{(1)!} = 0.00003

P(X = 2) = \frac{e^{-13}13^{2}}{(2)!} = 0.00019

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

For more than 15 days, the probability is:

P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)

Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

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