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ankoles [38]
2 years ago
14

An FBI survey shows that about 30% (i.e., 0.3) of all property crimes go solved. Suppose that in New York City 15 such crimes ar

e committed and they are each deemed independent of each other.
Mathematics
1 answer:
konstantin123 [22]2 years ago
7 0

Complete question :

An FBI survey shows that about 30% (i.e., 0.3) of all property crimes go solved. Suppose that in New York City 15 such crimes are committed and they are each deemed independent of each other.

a. What is the probability that exactly 3 of these 15 crimes will be solved? b. What is the probability that at most 3 of these 15 crimes will be solved? c. What is the probability that more than 11 of these 15 crimes will be solved?

Answer:

0.17

0.29686

0.000092

Step-by-step explanation:

Given that :

Probability of success (p) = 0.3

Number of cases (n) = 15

1 - p = 1 - 0.3 = 0.7

Usung binomial distribution formula :

a. What is the probability that exactly 3 of these 15 crimes will be solved?

P(x = 3)

Recall:

P(x = x) = nCx * p^x * (1 - p)^(n-x)

P(x = 3) = 15C3 * 0.3^3 * 0.7^12

P(x = 3) = 455 *

P(x = 3) = 0.17

B.) What is the probability that at most 3 of these 15 crimes will be solved?

P( X ≤ 3) = P(x = 0) + p(x = 1) + p(x = 2) + p(x = 3)

To save computation time, we can using the binomial probability calculator :

P( X ≤ 3) : 0.00474 + 0.03052 + 0.09156 + 0.17004 = 0.29686

c. What is the probability that more than 11 of these 15 crimes will be solved?

P( X > 11) = P(x = 12) + p(x = 13) + p(x = 14) + p(x = 15)

P( X > 11) = 0.000092

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Answer:

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Step-by-step explanation:

For each flight, there are only two possible outcomes. Either they arrive late, or they do not. The probability of a flight arriving late is independent of any other flight. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

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Show me how you solve it
julia-pushkina [17]

Answer:

5^20

Step-by-step explanation:

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>I</u>

\displaystyle \large{ \frac{ {a}^{m} }{ {a}^{n} }  =  {a}^{m - n} }

Therefore:

\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{8 - 3})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{5})^{4}  }

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>I</u><u>I</u>

\displaystyle \large{( {a}^{m} ) ^{n} =  {a}^{m \times n}  }

Thus:

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