Question

Partial Permutations

Find the number of 7-character (capital letter or digit) license plates possible if no character can repeat and: a) there are no further restrictions, b) the first 3 characters are letters and the last 4 are numbers, c) letters and numbers alternate, for example A3B9D7Q or 0Z3Q4A9

Combinations

A standard 52-card deck consists of 4 suits and 13 ranks. Find the number of 5-card hands where: a) any hand is allowed (namely the number of different hands) b) all five cards are of same suit c) all four suits are present d) all cards are of distinct ranks

Distribution Types

1) Which of the following sample spaces are uniform?

a) {land,sea} for a randomly point on a globe b) {odd, even} for a random integer from {1,2,. . . ,100} c) {leap year, non-leap year} for a random year before 2019 d) {two he{distance to origin} for a random point in {−3, −1, 1, 3} × {−4, −2, 2, 4} e) lads, two tails, one head and one tail} when flipping two fair coin

Inequalities

.in any uniform probability space: a) ?⊇? ⟶ ?(?)≥?(?) b) ?(?)≥?(?) ⟶ ?⊇? c) |?|≥|?| ⟶ ?(?)≥?(?) d) ?(?)≥?(?) ⟶ |?|≥|?|

Conditional Probability

Three fair coins are sequentially tossed. Find the probability that all are heads if: a) the first is tails b) the first is heads c) at least one is heads.

Answer 1

partial permutations

We know ,

the number of Capital letters = 26 (A-Z)

The number of digits are = 10 (0-9)

a) No restrictions :

We have to make 7 character plates then result will be

= 36×35×34×33×32×31×30

= 42072307200

b) first 3 are letters and last 4 are numbers

Then there are 26 letters and 10 numbers with out repetition gives as

= 26×25×24×10×9×8×7

= 78624000

c) Here letter and digit are permuted alternatively then there is two type , one starting with letter and other starting with digit and alternatively permuting the letter and digits then ,

No.of permutations starts with number ,

= 10×26×9×25×8×24×7

= 78624000

No.of permutations starts with letter ,

= 26×10×25×9×24×8×23

= 258336000

Then total number of permutations ,

= 258336000+78624000

= 336960000

Combinations

There are total of 52 cards

Number of suits = 4

Number of ranks =13 in each suit

a )

Any hand is allowed then we can select any 5 card from total of 52

Number of combination = 52C₅

=2598960

b)

we have to choose 1 suit and then any 5 cards from the suits having 13 rank in that

Number of combination ,

= 4C₁ × 13C₅

= 4 × 1287

= 5148

c)

all 4 suits have to be present then number of combination can be found as first selecting the suits we can select 4 suits , then selecting 5 cards from rank of each suits ,

Combination of selection of 5 cards ,

= 13C₁ ×13C₁ × 13C₁ ×13C₂

Here 2 hands are selected from 4th suit it can be from 1st , 2nd and 3rd so total combination is given as ,

= 4 { 13C₁×13C₁×13C₁× 13C₂)

= 4×171366

= 685464

d)

There are total of 13 ranks so the 5 selected cards should be of different ranks so number of selection is given as ,

1st card is selected from 1 of the 13 rank

2nd from an another rank then now available ranks are 12

So on to 5 cards

Then total combination is ,

= 13C₁ × 12C₁ × 11C₁ × 10C₁ × 9C₁

= 13×12×11×10×9

= 154440

Distribution types

The sample space is said to be uniform if each outcome of the sample space is equally likely.

a)

No

Given sample space is (land , sea ) in a globe the most part is covered with sea or water so the probability to get land is much less compared to sea so it's not a uniform sample space

b)

Yes

Between 1 to 100 there are 50 odd numbers and 50 even numbers with equal probability .

c)

No

Because the probability to get a leap year is very low compared to nonleap year because the leap year happens after 4 non leap years .

d)

No

Because let's consider any 2 points from the domain

1 ). ( -3 , -4 )

2 ). ( -1 , -2 )

The distance of these points from origin is calculated we get

1) .√ ( 9+16) = √ 25 = 5

2). √ (1+4) = √ 5

They are totally different , so they are not equally likely, so not uniform .

e)

No

When flipping two fair coins we get

{ (HH) , (HT) , (TH) ,( TT) }

Probability of two head = 1/4

Probability of 1 head and 1 tail = 2/4

Probability of two tails = 1/4

Not equal and hence not equally likely .

Conditional probability

​​​​​​If we toss three fair coin we get the sample space as

(HHH) , (HHT) , (HTH) , (THH) , (HTT) , (THT) , (TTH) , (TTT)

Conditional probability formula ,

P(A/B) =

a)

Probability of All are heads and the first is tail = 0

Answer = 0/ P( first is tail )

= 0

b)

Probability of All heads and the first is head = 1/8

Probability of first is head = 4/8

Answer  

=  

= 1/4

c)

Probability of all heads and atleast one is head = 1/8

Probability of atleast one is head = 7/8

Answer