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# APA Probability

Title

Name

41) (a) A Venn diagram

1. b) P(RuF) = P(R)+P(F)-P(R-F)

=29/50+23/50-8/50

=44/50

=0.88

1. c) 14+18+6=38

P(only one sport)

=38/50

= 0.76

1. d) P(Volleyball/Rugby)

=P(VuR)/P(R)

=9/29

=0.31

1. e) P(Field Hockey/Not Rugby)=15/21

=5/7

=0.714

35) (a) P(A’) and P(B’)

=7/100x13/100

=91/10000

(b) P(A) and P(B)

=93/100x87/100

=8091/10000

(c) At least one of the test detects steroid = 1-None shows presence of steroid

= 1-91/10000

=9909/10000

6) a) We have to fill 4 spaces. We are left with 8 numbers to choose from (1,2,3,4,5,6,8,9)

4P4*7

7*5 ways – 5P48*7.

To eliminate zero, we fix zero in the first and rearrange the remaining number in the last four spaces. Hence, 4P4*7 such cases,

The total codes = 5P4*8*7)-4P4*7

=280-28

=252 codes

1. b) 3 as the third and 5 as the fifth=5!/(n-2)!

=5x4x3x2x1/(5-2)!

=120/6

=20 Codes

1. c) To begin with an even number, The 1st letter can be arranged in 4! Different forms starting with the four even numbers, that is 2,4,6,8

The 2nd,3rd, 4th and 5th can be arranged in 4! Ways. Hence, number of arrangement = 4!x4!

=576codes.

1. d) Begin and end with even number.

1st – 4!

5th – 4!

2nd, 4th and 5th can be arranged into 3! ways.

Hence, The total codes=4!x4!x3!

=576x6

=3456 codes.

23) Let 5 be the 1st digit. The other four digits can have either of the other 9 choices, as we have excluded 8. This gives 9^4 numbers.

Let 5 be the 2nd digit. The first digit can have either of 7 choices, excluding 5, 8 and 0. The other three digits can have any of the 9 choices. This gives 7 * 9^3 numbers.

Let 5 be the 3rd digit. The first digit can have either of 7 choices, excluding 5, 8 and 0. The 2nd digit can have 8 choices excluding 5 and 8. The other two digits can have any of the 9 choices. This gives 7 * 8 *9^2 numbers.

Let 5 be the 4th digit. The first digit can have either of 7 choices, excluding 5, 8 and 0. The 2nd and 3rd digits can have any of 8 choices. This gives 7 * 8^2*9 numbers.

Let 5 be the 5th digit. The first digit can have either of 7 choices, excluding 5, 8 and 0. The 2nd, 3rd and 4th digits can have any of 8 choices. This gives 7 * 8^3 numbers.

Hence, the total possible numbers;

= 9^4 + 7*9^3 + 7*8*9^2 + 7*8^2*9 + 7*8^3= 23816

## Cite This work.

Urgent Homework (2022) . Retrive from https://www.urgenthomework.com/sample-homework/apa-probability

"." Urgent Homework ,2022, https://www.urgenthomework.com/sample-homework/apa-probability

Urgent Homework (2022) . Available from: https://www.urgenthomework.com/sample-homework/apa-probability

[Accessed 18/08/2022].

Urgent Homework . ''(Urgent Homework ,2022) https://www.urgenthomework.com/sample-homework/apa-probability accessed 18/08/2022.

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