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Harlamova29_29 [7]
3 years ago
14

Investing is a game of chance. Suppose there is a 36% chance that a risky stock investment will end up in a total loss of your i

nvestment. Because the rewards are so high, you decide to invest in five independent risky stocks. Find the probability that at least one of your five investments becomes a total loss. Round to the nearest ten-thousandth when necessary.
Mathematics
1 answer:
Sergio039 [100]3 years ago
3 0

Answer: P(x ≥ 1) = 0.893

Step-by-step explanation:

We would assume a binomial distribution for the outcome of the investment. The formula is expressed as

P(x = r) = nCr × p^r × q^(n - r)

Where

x represent the number of successes.

p represents the probability of success.

q = (1 - r) represents the probability of failure.

n represents the number of trials or sample.

From the information given,

p = 36% = 36/100 = 0.36

q = 1 - p = 1 - 0.36

q = 0.64

n = 5

Therefore,

P(x ≥ 1) = 1 - P(x = 0)

P(x = 0) = 5C0 × 0.36^0 × 0.64^(5 - 0)

P(x = 0) = 1 × 1 × 0.107

P(x = 0) = 0.107

P(x ≥ 1) = 1 - 0.107 = 0.893

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Answer:

P(X \geq 1) = 1-P(X

And we can find the individual probability like this:

P(X=0) = 16C0 (0.5)^0 (1-0.5)^{16-0} = 0.0000153

And replacing we got:

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Step-by-step explanation:

Assuming the following question: With the salary cap in the NFL, it is said that on any given Sunday any team could beat any other team. If we assume every week of the 16 week season a team has a 50% chance of winning, what is the probability that a team will have at least 1 win?

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem

Let X the random variable of interest, on this case we now that:  

X \sim Binom(n=16, p=0.5)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

And we want this probability:

P(X \geq 1)

And using the complement rule we got:

P(X \geq 1) = 1-P(X

And we can find the individual probability like this:

P(X=0) = 16C0 (0.5)^0 (1-0.5)^{16-0} = 0.0000153

And replacing we got:

P(X \geq 1) = 1-P(X

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