15/10/2018 7:59 pm
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PnC , Probability
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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?
26 questions & discussions are there under this sub-topic
2
03/01/2024 2:00 pm
A bag I contains 5 red balls and 3 green balls, bag II contains 8 red balls and 4 green balls and bag III contains 5 red balls and 5 green balls. A ball is picked randomly from a bag and it is found to red. Find the probability that the ball is drawn from bag I ?
1
15/10/2018 10:28 pm
what should be the sample space in question 45?
aniket prajapati 15/10/2018 10:59 pm
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
0
15/10/2018 10:26 pm
solution for question 46 please.
0
15/10/2018 10:29 pm
what should be the sample space in question 45?
0
15/10/2018 10:31 pm
solution please
aniket prajapati 15/10/2018 11:19 pm
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
0
15/10/2018 10:33 pm
solutions for question 31,32 and 33 please
aniket prajapati 15/10/2018 11:42 pm
This post was modified 6 years ago by aniket prajapati
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
aniket prajapati 15/10/2018 11:49 pm
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
0
16/10/2018 11:17 am
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.
aniket prajapati 16/10/2018 10:22 pm
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
0
17/10/2018 12:50 am
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.
0
17/10/2018 11:34 pm
Each of two persons tosses 3 fair coins . The probability that they obtain the same number of heads is ?
Utkarsh Garg 17/10/2018 11:49 pm
This post was modified 6 years ago by Utkarsh Garg
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
0
17/10/2018 11:36 pm
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 ?
Utkarsh Garg 17/10/2018 11:52 pm
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
0
17/10/2018 11:38 pm
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 ?
0
19/10/2018 10:36 am
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?
TG Team Topic starter
19/10/2018 12:19 pm
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
0
25/10/2018 8:23 pm
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