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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?

20 Answers

EB8E12EB DF67 44D4 8950 F84CBE1A5EB1

solution for question 48 please


Assuming the seating assignments are fixed, the problem is tantamount to arranging 17 blue chairs and 33 green chairs in a circle so that no two of the green chairs are adjacent. We will solve the problem for a line, then adjust our answer because chairs are arranged in a circle.

Line up 17 blue chairs. This creates 18 spaces, 16between successive blue chairs and two at the ends of the row. Choose three of the spaces in which to place a green chair. This can be done in (18C3)ways.

However, we have counted selections in which both the first and last chairs are green. If both ends are selected, there are 16 remaining spaces in which we place the third green chair. Hence, there are 16 linear arrangements in which no two of the green chairs are adjacent in which there is a green chair at both ends of the row. Since we will be joining the ends of the row to form a circle, these arrangements must be excluded.

Therefore, the number of ways of circular arrangements of 17 blue chairs and 33 green chairs so that no two of the green chairs are adjacent is 

18C3 - 16C1= 3*17*16-16 =16*50=800 (answer)

read 33 as 3 ,sorry for the mistake


5B7E5685 9117 4874 976B 46A27A15D402

solution for question 46 please.


7C5C59F8 1699 48AD 90FE EE9ECFA74253

what should be the sample space in question 45?


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



sample space should be 4!


7C5C59F8 1699 48AD 90FE EE9ECFA74253

what should be the sample space in question 45?


9019730B E3E7 4793 A30A 00953E30FB39

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


hence answer is 3


375C77AC 0363 4A51 9CDA C984E1B60C74

solutions for question 31,32 and 33 please


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 numbers and we can choose any numbers from the remaining 16 numbers as



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



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!



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?


         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)



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 

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

Cubes painted on three sides : 8 
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 




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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. 

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