Tickets for movies in a particular theatre have numbers from 1 to 100 printed on them with each ticket having a unique number. The tickets are sold in a random order. Three people go to the theatre and buy tickets, when all the tickets are available. What is the probability that exactly one person's seat is in the range of 6 to 50?

what should be the sample space in question 45?

45th

total 4 digit no. formed =4!

it will be divisible by 4 only when the last two digits will be 28,24,48,32,84

total no. formed when it ends with 28=2*1*1*1=2

similarly every case will have 2 nos.

total 5*2=10

probability=10/4!=5/12

sample space should be 4!

solution for question 46 please.

what should be the sample space in question 45?

solution please

for first cube: 5 black (B1) faces and 1 white (W1)

probability of getting black on top= P (B1)=5/6

probability getting white on top= P(W1)=1/6

For second face let number of black faces be x, hence number of white faces will be 6-x as cube as 6 faces.

Probability of getting black on top= P(B2)=x/6

Probability of getting white on top=P(w2)=6-x/6

same color will be obtained if both faces are black or both faces are white

probability of getting same color=1/2

P(B1) * P(B2)+P(W1)* P(W2)=1/2

(5/6)*(x/6) +(1/6){(6-x)/6}=1/2

x=3

hence answer is 3

solutions for question 31,32 and 33 please

33

We start choosing by the lowest number, and will not choose any number before that number and I proceed by choosing only in increasing order.

So, when we choose the lowest number, we eliminate the number next to the chosen number. We continue this process and on the final selection of number, We have chosen all 10 numbers so we do not have to eliminate any number anymore.

If one thinks carefully, then one will observe that we have eliminated 9 numbers and we can choose any numbers from the remaining 16 numbers as

16C10

31st

Since there are six faces, you can form 6*4=24 right-angled triangles on the faces.

and 24 diagonally so total right anled triangles formed=48

total triangles formed by cube=8C3=56

probability=48/56=6/7

In how many ways can we form five digit number composed of 1, 2, 3, 4 and 5 exactly once such that 3 always follows 2 and 4 always precedes 5 but 1 doesn’t precede 2? For example: 21435 is a valid number but 12345 is not.

total nos, =5!

group 23 as single digit and 45 as single digit now we have 3 digit as (23)(45)(1)

no. formed with such arrrangement =3!

probability=3/5

A gardener plants three mango trees, four orange trees, and five banana trees in a row. Find the probability that no two banana trees are next to one another.

solution please..

Each of two persons tosses 3 fair coins . The probability that they obtain the same number of heads is ?

is it 5/16?

S = { HHH, HHT, HTH, THH, TTH,

THT, HTT, TTT }

So, n(S) = 8.

But both A & B are tossing coins.

So, in our case, n (S) = 8^2 = 64.

Out of 8 sample points in a toss of 3 coins, 3 have 2 heads, 3 have 1 head, 1 has 3 heads & 1 has 0 heads.

Again, as tossing is performed by both A & B, firstly, we've to square the results obtained & then add them.

n (A) = 3^2 + 3^2 + 1^2 + 1^2

**=** 20.

Probability = 20/64 = 5/16

It is known that at noon at a certain place the sun is hidden by clouds on an average 2 days out of every 3 . The chance that the sun will be shining at noon on at least 4 out of 5 specified Future days is ?

P(shining)= 1/3 P(not shining)= 2/3

P(that the sun will be shining at noon on at least 4 out of 5 specified Future day)=

(1/3)^5 + (1/3)^4* 5c1(2/3)

11/243

The odds in favour of A winning a game of chess against B are 5: 2. If 3 games are to be played ,then the odds in favour of A's winning at least one game are ?

All the faces of a 4 × 4 × 4 cube are painted and then the cube is cut into 64 unit cubes . If one of the cubes is selected and rolled what is the probability that out of the five visible faces two our painted?

Cubes painted on two side : 24

2 faces are colored, so probability that the cubes lands on un-colored face is 4/6

Cubes painted on three sides : 8

3 faces are colored, so probability that the cubes lands on un-colored face is 3/6

Probability : (24 x 4/6 + 8 x 3/6 )/ 64 = 5/16

Please help me with the approach for both the questions.

The probability of 0 hits out of n is nCo × (1/2)^n and 1 hit out of n is nC1 × ( 1/2)^n

Probability of destroying > 0.99...... = 1-{nC0×(1/2)^n + nC1×(1/2)^n } > 0.99

=> 1 - ( n +1)/2^n > 0.99

=> 1- 0.99 > ( n +1)/2^n

Smallest n = 11.

each of 2 persons tosses three fair coins. The probability that they obtain the same no of heads is?

number of possible outcomes when 3 fair coins are tossed => 2^3

{ TTT , TTH , THH , HHH , HHT , HTT , HTH , THT}

Probability of A getting 3 heads => 1/8

So probability that both will get 3 heads is 1/8 x 1/8 = 1/64

Probability of A getting 2 heads => 3/8

so probabilty that bothwill get 2 heads => 3/8 x 3/8

Similarly probability of getting 1 heads => 3/8 x 3/8

and probabilty of getting o heads => 1/8 x 1/8

2/64 + 18/64 = 5/16

but the ans is 5/16

1/64 + 1/64 + 9/64 +9/64 = 5/16.