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### General Statistics probability calculations

General Statistics

Name

Affiliation

Choosing an Outfit

3 pairs of pants (B, R, O)

2 tops (W, R)

2 shoes (B, W)

a.Tree Diagram

# of outfits = 3 ´ 2 ´ 2 = 12

2.How many 3-digit numbers can be formed if:

a. there are no restrictions? ANS: 9 ´ 10 ´ 10 = 900 (a 3 digit number can’t start with zero)

b.there are no repetitions? ANS: 9 ´ 9 ´ 8 = 648 (gain the zero, lose the 1st digit number)

3.How many 4-letter arrangements can be formed if:(26 letters in the alphabet?)

a.there are no repetitions? ANS: 26 ´ 25 ´ 24 ´ 23 = 358 800

b.repetitions allowed? ANS: 264 = 456 976 a.k.a. 26 x 26 x 26 x 26 =

4.How many numbers can be formed from the digits 1, 4, 6, and 8 (one of each) if the number:

a. Must be even ANS: 3 ´ 2 ´ 1 ´ 3 (do first) = 18

b. Must be greater than 5 000. ANS: 2 ´ 3 ´ 2 ´ 1 = 12

C. Must be greater than 400. ANS: 3 ´ 3 ´ 2 + 4 ´ 3 ´ 2 ´ 1 = 18 + 24 = 42

1.How many 5-letter arrangements can be made from the letters A, B, C, D, E, F, G?

ANS: 7P5 or 7 ´ 6 ´ 5 ´ 4 ´ 3 = 2 520

2.How many ways can 6 books be arranged on a shelf?

ANS: 6P6 or 6 ´ 5 ´ 4 ´ 3 ´ 2 ´ 1 = 720

3.5-letter arrangements are formed from the letters in NUMBERS. How many if:

a.no vowels? ANS: 5 ´ 4 ´ 3 ´ 2 ´ 1 = 120

b.1st and last letters are consonants? ANS: 5 ´ 5 ´ 4 ´ 3 ´ 4 = 1 200

___ x ___ x ___ x ___ x ____AEIOU are vowels in the alphabet

Con Any Any Any Con

4. 6-letter arrangements are formed from ABCDEF. How many if:

no restrictions?

ANS: 6 ´ 5 ´ 4 ´ 3 ´ 2 ´ 1 = 6! = 720

b.A & B together? (think of AB as one letter)

ANS: 5 ´ 4 ´ 3 ´ 2 ´ 1 ´ 2 ´ 1 = 120

Number of ways to arrange AB ¿

The 2 ´ 1 is the placing of the A and B

The 5 ´ 4 ´ 3 ´ 2 ´ 1 are the placing of the AB as one unit and the CDEF as 4 units.

c.A, B, & C together? ANS: 4 ´ 3 ´ 2 ´ 1 ´ 3 ´ 2 ´ 1 = 144

The 4 ´ 3 ´ 2 ´ 1 is the arranging of DEF (in any order) and the ABC as a whole.

The 3 ´ 2 ´ 1 is the arranging of the ABC.

d.A & B separated? (arrange others first) ANS: 4 ´ 3 ´ 2 ´ 1 ´ 5 ´ 4 = 480 OR #a – #b

The 4 ´ 3 ´ 2 ´ 1 is the CDEF

The 5 is the A or B (one or the other) and the CDEF

You basically have __ C __ D __ E __ F ___

Now the CDEF can go in any order hence the 4!

The A or B could go in any of the 5 spaces hence the multiplying 5

Once the A or B is placed, the other one has 4 difference places to go, hence the multiplying 4.

e.A, B, & C separated? ANS: 3 ´ 2 ´ 1 ´ 4 ´ 3 ´ 2 = 144 (can’t do as above)

Same hint as above. You basically have __ D __ E __ F ___

Now the DEF can go in any order hence the 4!

The A or B or C could go in any of the 4 spaces hence the multiplying 4

Once the A B or C is placed, the other two have 3 difference places to go, hence the multiplying 3.

The multiplying 2 is the last A, B or C to go in either of the empty two spots.

f.the word BAD appears? ANS: 4 ´ 3 ´ 2 ´ 1 = 24

The 4 is the word BAD or C, E, or F (4 items as BAD is one unit)

The 3 is the 4 options above less the one already used.

The 2 is the 3 options above less the one already used.

The 1 is the only option left.

5. How many arrangements of the letters in the word:

BANANA

ANS: EMBED Equation.DSMT4 3 A’s and 2 N’s

STATISTICS

ANS: EMBED Equation.DSMT4 3 S’s, 3 T’s and 2 I’s

Computation

A committee of 3 is selected from 5 girls and 7 boys. How many if:

a.no restrictions? ANS: 12C3 = 220

b.all boys? ANS: 7C3 = 35

c.1 girl and 2 boys? ANS: 5C1 · 7C2 = 105

d.At least 1 girl? ANS: 1 girl or 2 girls or 3 girls = 5C1 · 7C2 + 5C2 · 7C1 + 5C3

= 105 + 70 + 10

= 185

OR #a – #b = 220 – 35

= 185

2. 10 points are on a circle (no 3 are collinear). How many:

Different lines are determined?

ANS: 10C2 = 45

781685152400A line is a connection of 2 points. So, take the 10 points and connect 2 of them.

ANS: 10C4 = 210

A quadrilateral has 4 sides. So, take 10 points and connect 4 of them.

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