Test: Probability- 2 - Computer Science Engineering (CSE) MCQ

There are only two possible sets whose elements sum to 22 : {6,6,6,4}, {6,6,5,5}
Number of permutations for 1st set : 4!/3! = 4
Number of permutations for 2nd set : 4!/(2!*2!) = 6
So total number of ways to sum 22 = 10
So X =10.

It immediately follows from the monotonicity property that ,
0 <= P(E) <= 1 ,
probability of atleast one means union of the probability of events , i.e.
P(A U B) = P(A) + P(B) - P(A⋂B) 
here , P(AUB) = 1 , because it can not be more than 1 and if atleast one of event have probability 1 (here , P(A) = 1) , then union of both should be 1 .
so , P(A U B) = P(A) + P(B) - P(A⋂B)
1 = 1 + 1/2 - P(A⋂B) ,
P(A⋂B) = 1/2 , 
now , P(A/B) = P(A⋂B) / P(B) = (1/2) / (1/2) = 1 ,
P(B/A) = P(A⋂B) / P(A) = (1/2) / (1) = 1/2 . 
hence option (D) , 
NOTE :- if atleast one of the two events has/have the probability 1 . then both events should be independent events but vise - versa not true . 

Let the probability of winning against player A be  and the probability of winning against player B be b.

 be the probability of winning the series in which the games played are against x,y and z in order.

We can see that not all probabilities are equal, so option E is not correct.
We can also see that options A and D result in the same value, so they are not correct either.
Comparing option B and option C.

Hence, option B is the correct answer.

Case 1:largest is 5 , smallest 1 and middle is 2 or 3 or 4 : 3*3!
Case 2.largest is 5 , smallest 1 and middle is 1 or 5 : 3!*2/2!
Case 3:largest is 6 , smallest 2 and middle is 3 or 4 or 5 : 3*3!
Case 4:largest is 6 , smallest 2 and middle is 6 or 2: 3!*2/2!
So probability the highest and the lowest value differ by 4 =( 3*3!+3!*2/2!+3*3!+3!*2/2!)/6³ =2/9

B) for all the 6 different faces 1,2,3,4,5,6 the probability is
1/6⁶ and then for
6! different permutations :
6! / 6⁶

Here we are having a total of 1000 Balls, out of which we firstly draw 100 balls , and then 101^st ball..
Firstly we have to find expected number of white and black balls in drawn 100 balls , as both can occur in 100 balls..
We have a situation like this:

Expected number of white balls = n*W/N = 100*(100/1000) = 10
Expected number of black balls = n*B/N = 100*(900/1000) = 90
So,  we have drawn 100 balls(90 black, 10 white)
Left balls = (810 Black , 90 white) = 900 total
Now,

probability for 101^st ball to be black = 810/900 = 9/10
So, option (A) is Correct 

there n binary bits that can differ but only d should differ in this case
ways of choosing these d bits = ⁿC_d probability of d bits differ but, n - d bits do not differ = (1/2)^d(1/2)^{(n - d)}
no of ways = ⁿC_d/2ⁿ

probability of obtaining an even number of heads in 5 tosses, zero being treated as an even number number of event = 0 head or 2 head or 4 head
Probability of head = 2/3
Probability of tail = 1/3
Probability = 
= 121/243

Tourists are 2/3 of population of the park
Among them probability of correct answer is 3/4
so, 2/3*3/4=1/2

As, %of Doctor A successful for stage III 78/80*100=97.5
for stage IV 2/20*100=10
As, %of Doctor B successful for stage III 49/50*100=98
stage IV 11/50*100=22

Even number = 2,4,6
odd number = 1,3,5
Product will come even when even one time even number comes odd product will be if every time odd comes 

out of 4 times 2 times head should be present
ways of selecting these 2 places 4C2 probability of getting 2 heads and 2 tails = (1/2²)(1/2²)
probability = ⁴C₂/2⁴ = 3/8

P{3,5} = 1 - P{2,4,6} - P{1} = 1/2 - 1/6 = 1/3
Can P{3} = 0? then P{6} = 1/3 and P{2,4} = 1/2 - 1/3 = 1/6. And P{5} = 1/3. Possible. So, option D.

probability of getting all heads = 1/16
probability of getting all tails = 1/16
probability of getting at least one head and one tail = 1 - 1/16 - 1/16 = 7/8

Let the 4 tosses be named P,Q,R and S
To have 3 consecutive heads:

Thus, the probability of getting 3 consecutive heads is given by:

Hence, option D is the correct answer.
Another way of looking at it is:

We know that the sum of all the probabilities is 1.
Therefore on integrating    with limits a to 1, the result should be 1.

Hence a = 0.5

Integrate 1/x² in the limits a to 1.. and equate it to 1.. solving we get a=0.5

1st time it is 0.25(1/4), when tail tail comes, entire process gets repeated, so next time probability of Y to happen is 0.25*0.25 ((1/4)*(1/4)), likewise it goes on as infinite GP
Sum of infinite GP = a/(1-r)
here a= 1/4 and r = 1/4
so answer becomes 1/3 i.e 0.33

Answer should be 0.33 P(TH)=1/4
P(HH + HT)=1/2
now if TT comes then toss again,
So, P(TTTH)= 1/16 . and so on.... P(TH+TTTH+........) = 1/4 + 1/16+... = 1/3

Given that probability of getting 2 or  4 or 6 is same.

Given that A man has three cats and  At least one is male.
Possible combination for atleast one cat is male=(M,F,F),(F,M,F),(F,F,M),(M,F,M),(M,M,F),(F,M,M),(M,M,M)

Probability that all three are male = 1/7
Hence,Option(B)
1/7 is the correct Choice.

First digit can be chosen in 8 ways from 1−9 excluding 7
Second digit can be chosen in 9 ways from 0−9 excluding 7 and similarly the third digit in 9 ways.
So, total no. of ways excluding 7 = 8∗9∗9
Total no. of ways including 7 = 9∗10∗10
So, ans = (8∗9∗9)/(9∗10∗10)=18/25

There are  12 houses on one side of a street are numbered with even numbers.
In which 5 houses are strictly greater than Number 14.
And remaining 7 houses are numbered smaller than 14 (i.e. including 14)
No of way of choosing at least 2 of these newspapers are delivered to houses with numbers strictly greater than 14.
5C₃  +5C₂ * 7C₁ =80
Total way of choosing 3 houses=12C₃ =220
So Required probability=80/220=4/11

If we want lime juice then we need to throw {2 or 6} . And If we don't get {2 or 6} in first go, then we need to throw {1 or 3 or 5} in first go and again we will try to get {2 or 6} in 2nd throw and so on.
Note - If we throw 4 in any go then we will end up getting lassi. But we want lime juice.
So probability of getting lime juice = (2/6)  + (3/6) * (2/6) + (3/6)2 * (2/6) + .......  to infinity.  

here i check option---> i let P = probability,  P and Q is replaced by A and B.
A) it  is saying independent means P(A ^ B) =  P(A)* P(B)  so option A is false
B) P(AÜ B) =  P(A) + P(B) - P(A ^ B)
hence P(AÜB) >= P(A) + P(B)  its false option B is false
C) if A and B is mutually exclusive means P(A ^ B) =  0 then its independent ........its false  there is no relation independent and mutually exclusive
D)  P(A ^ B) <= P(A)

its true

hence option D is correct

for every car accident we can pick a day in 7 ways
total number of ways in which accidents can be assigned to days = 7⁷
probability of accidents happening on a particular day = 1/7⁷
we can choose a day in 7 ways
hence probability = 7/7⁷= 1/7⁶

let probability of Event E1 = x = prob of E2
prob(E1 union E2) = prob(E1) + prob(E2) - prob(E1 intersect E2)
1 = x + x -x² (prob(E1 intersect E2) = prob(E1) * prob(E2) as events are independent)
x = 1

There are 4 sets of cards. So, up till 8 cards there is a chance that no more than 2 cards are from a given set. But, once we pick the 9^th one, it should make  3 cards from any one of the sets. So, (C) is the answer.

probability of first ball white and second one black = (10/25)x(15/24)
probability of first ball black and second one white = (15/25)x(10/24)
probabilty = sum of above two probabilities = 1/2

no of integers divisible by 2 = 50
no of integers divisible by 3 = 33
no of integers divisible by 5 = 20
no of integers divisible by 2 and 3 = 16
no of integers divisible by 2 and 5 = 10
no of integers divisible by 3 and 5 = 6
no of integers divisible by 2 and 3 and 5 = 3
total numbers divisible by 2 or 3 or 5 = 50 + 33 + 20 -16 -10 - 6 + 3 = 74
total number not divisible by 2 or 3 or 5 = 26
probability = 0.26 [EDIT]

prob(at least one head) = 3/4
prob(both heads) = 1/4
using bayes' theorem = (1/4)/(3/4) = 1/3