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Tems11 [23]
3 years ago
7

For a group of 50 people, assuming that each person is equally likely to have a birthday on each of 365 days in the year, comput

e
(a) The expected number of days of the year that are birthdays of exactly 3 people:
Mathematics
1 answer:
kramer3 years ago
6 0

Answer:

0.0177 days

Step-by-step explanation:

Solution:-

- The group has n = 50 people

- Each person is equally likely to have a birthday on each of 365 days in the year.

- The probability of success p = 1 / 365

- The probability of failure q = (1 - 1/365)

- We will denote random variable X as a given day has k number of birthdays.

- X follows binomial distribution:

                            X ~ Bi ( 50 , 1/365)

- The probability that there would be 3 birthdays on a day would be:

                           P ( X = 3 ) = 50C3 ( 1 / 365)^3 * ( 1 - 1/365)^47

                           P ( X = 3 ) = 0.00035

- The expected number of days for exactly three people to have same birth date:

                          E( X = 3 ) = P (X=3)*n = 0.00035*50 = 0.0177 days

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<h3>(2x - 4) - (6x + 6) = -2( 2x + 5 )</h3>

<h3>Further explanation</h3>

Order of Operations in Mathematics follow this following rule :

  1. Parentheses
  2. Exponents
  3. Multiplication and Division
  4. Addition and Subtraction

This rule is known as the PEMDAS method.

Let us tackle the problem!

(2x - 4) - (6x +6) = 2x - 4 - 6x - 6 ⇒ open parenthesis

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(2x - 4) - (6x +6) = (2 - 6)x - 10 ⇒ factorize

(2x - 4) - (6x +6) = -4x - 10

\large {\boxed {(2x - 4) - (6x +6) = -2( 2x + 5 ) } }

<h2>Another Example :</h2>

(2x - 4) + (6x +6) = 2x - 4 + 6x + 6 ⇒ open parenthesis

(2x - 4) + (6x +6) = 2x + 6x - 4 + 6 ⇒ rearrange term

(2x - 4) + (6x +6) = (2 + 6)x + 2 ⇒ factorize

(2x - 4) + (6x +6) = 8x + 2

\large {\boxed {(2x - 4) + (6x +6) = 2( 4x + 1 ) } }

<h3>Learn more</h3>
  • Infinite Number of Solutions : brainly.com/question/5450548
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<h3>Answer details</h3>

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