GATE Computer Science Engineering(CSE) 2027 Test: Probability- 1 Free Online

Here our sample space consists of 3 + 3 * 6 = 21 events- (4), (5), (6), (1,1), (1,2) ... (3,6).
Favorable cases = (6), (1,5), (1,6), (2, 4), (2, 5), (2, 6), (3, 3), (3,4), (3,5), (3,6)
Required Probability = No. of favorable cases/Total cases = 10/21
But this is wrong way of doing. Because due to 2 tosses for some and 1 for some, individual probabilities are not the same. i.e., while (6) has 1/6 probability of occurrence, (1,5) has only 1/36 probability. So, our required probability 
= 1/6 + (9 * 1/36) = 5/12

The probability of selecting a red ball  = (1/3) * (2/5) + (2/3) * (3/4) = 2/15 + 1/2 
 = 19/30 
Probability of selecting a red ball from box P = (1/3) * (2/5) = 2/15 
Given that a ball selected in the above process is a red ball, the probability that it came from the box P is = (2/15) / (19/30) = 4/19

10% email are spam, i.e. 90% email are not spam 90% of mail marked as spam is spam, 10% mail marked as spam are not spam By Bayes theorem the probability that a mail marked spam is really a spam  

multiply both side with  and subtract:

ways of getting n heads out of 2n tries = ²ⁿC_n
probability of getting exactly n heads and n tails = (1/2ⁿ)(1/2ⁿ) number of ways = ²ⁿC_n/4ⁿ

table has 3 coins with H,H,T facing up.
now, probability of chosing any coin is 1/3 , as we can chose any of three coins.
Case A: 1st coin :  either H or T can come.
so, HHT THT possible.only HHT is favourable.
which gives (1/3)*(1/2)= 1/6
Case B: 2nd coin :  either H or T can come.
so, HHT HTT possible.only HHT is favourable.
which gives (1/3)*(1/2)= 1/6
case C: 3rd coin : Table already contains two H's so, whatever comes is favourable.
which gives (1/3)*1 = 1/3
summing up total gives 1/6 + 1/6 + 1/3 = 2/3

suppose there are k places within n bit string where mismatch has occoured probability of this occouring is ⁿC_k(prob. of mismatch)k(prob. of match)^{(n - k)} = ⁿC_k(1/2)^k(1/2)^(n-k) = ⁿC_k(1/2)ⁿ
k can range from 1 to n hence required probability sum(ⁿC_k(1/2)ⁿ) where k ranges from 1 to n
hence (1/2ⁿ)(2ⁿ - 1)

Alternatively Probability of matching at given place 1/2
there are n places hence probability of matching 1/(2ⁿ)
hence probability of mismatch 1 - 1/(2n)

Probability of India win =1/2
Probability of India lost=1/2
If really India wins, then Karan lies i.e.=2/3 and Arjun tells truth=3/4
Now prob. of Karan lies  and Arjun tells truth=2/3 * 3/4=1/2
Now prob. of Arjun lies  and Karan tells truth=1/4 * 1/3=1/12
so, by Bayes theorem 

Suppose that a shop has an equal number of LED bulbs of two different types.  ==> Therefore
Probability of Taking Type 1 Bulb => 0.5
Probability of Taking Type 2 Bulb => 0.5
The probability of an LED bulb lasting more than 100 hours given that it is of Type 1 is 0.7, and given that it is of Type 2 is 0.4.
Prob(100+ | Type1) => 0.7
Prob(100+| Type2) => 0.4
Prob(100+) => Prob(100+ | Type1) * Prob(Type1) + Prob(100+ | Type2) * Prob(Type2)              
= 0.7 * .5 + .4 * .5
= 0.55

SIMPLY-
TYPE-1 bulbs ===>let total =10 then 7 are going for 100 years
TYPE-2 bulbs====>let total=10 then 4 are going for 100 years
TOTAL==>20 out of which 11 are going for 100 years ..THEREFORE probability=11/20=.55

Let,
P = P applies for the job
Q = Q applies for the job
 

(a) is true only if events are independent.
(b) is true only if events are mutually exclusive i.e.
(c) is false everywhere
(d) is always true as  P(A U B) = P(A) + P(B) — P(A ∩ B)

Each of the nine words have equal probability. So, expected length

Probability of choosing the correct option = 1/4
Probability of choosing a wrong option = 1/3
So, expected mark for a question for a student 
Expected mark for a student for 150 questions 
So, sum total of the expected marks obtained by all 1000 students

Let the expected number of coin flips be . The case analysis goes as follows:
a. If the first flip is a tails, then we have wasted one flip. The probability of this event is 1/2 and the total number of flips required is X + 1.

b. If the first flip is a heads and second flip is a tails, then we have wasted two flips. The probability of this event is 1/4 and the total number of flips required is X + 2 as the same scenario as beginning is there even after 2 tosses.
c. If the first flip is a heads and second flip is also heads, then we are done. The probability of this event is 1/4 and the total number of flips required is 2.

Thus, the expected number of coin flips for getting two consecutive heads is 6.  

using static huffman compression you can encode the more common colours in fewer bits than the rare colours, that being the case on can expect that common colours will usually be chosen.
eg:
red    1
blue   01
green  001
white  0001
black  0000

on average from 16 draws there will be

8 reds   = 8 bits
4 blues  = 8 bits
2 greens = 6 bits
1 white  = 4 bits
1 black  = 4 bits

for a total of 30/16 = 15/8 bits on average

total outcomes = 36
expected outcomes={(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)}=15
hence Probability=15/36=5/12

= 0.41666

In Poisson distribution :
Mean  =  Variance  as  n is large and p is small
And we know :

Hence 54 should be the right answer..

20 request in 1 hour.. so we can expect 15 request in 45 minutes...
So, lemda = 15.. (expected value)
poission distribution formula: f(x, lemda) = p(X = x) = (lemda ^ x * e ^ - lemda)  / x!
Therefore p(one request) + p(3 request) + p(5 request)
= p(1; 15) + p(3; 15) + p(5; 15)
= 6.9 * 10^3 * e ^ -15..
= 6.9*10^3*e^-15 = 6.9*10^3*e^5*e^-20 = 1.02*10^6*e^-20..  Ans is (B)

Poisson Probability Density Function (with mean 
We have to sum the probability density function for  (thus finding the cumulative mass function)