UGC NET CS 2011 DecPaper2
Question 1 
Which of the following data structure is Nonlinear type ?
Strings  
Lists  
Stacks  
None of the above 
Question 1 Explanation:
A data structure is said to be linear if its elements form a sequence or a linear list.
Examples:
1. Array
2. Linked List
3. Stacks
4. Queues
Examples:
1. Array
2. Linked List
3. Stacks
4. Queues
Question 2 
The total number of comparisons in a bubble sort is
0(log n)  
0(n log n)  
0(n)  
None of the above 
Question 2 Explanation:
The total number of comparisons in a bubble sort is O(n^{2})
Question 3 
Which of the following is a bad example of recursion ?
Factorial  
Fibonacci numbers  
Tower of Hanoi  
Tree traversal 
Question 3 Explanation:
→ Fibonacci Series using recursion using condition is
Condition: fibonacci(n1) + fibonacci(n2)
Recurrence relation is: T(n)=2T(n1)+C
Time Complexity: O(2^{n})
→ Factorial number using recursion is
Condition: fact(n1)*n
Recurrence relation is: T(n)=T(n+1)+1
Time Complexity: O(n)
→ Tower of Hanoi using recursion is
Step1: Hanoi(disk  1, source, aux, dest)
Step2: move disk from source to dest
Step3: Hanoi(disk  1, aux, dest, source)
Recurrence relation is: T(n)=2T(n1)+C
Time Complexity: O(2^{n})
→ Tree traversals like Preorder,Postorder and Inorder using recursion only
Note: Recursion is best in the case of Tower of Hanoi,Factorial number and Tree traversals but it is not give good performance to fibonacci numbers.
Condition: fibonacci(n1) + fibonacci(n2)
Recurrence relation is: T(n)=2T(n1)+C
Time Complexity: O(2^{n})
→ Factorial number using recursion is
Condition: fact(n1)*n
Recurrence relation is: T(n)=T(n+1)+1
Time Complexity: O(n)
→ Tower of Hanoi using recursion is
Step1: Hanoi(disk  1, source, aux, dest)
Step2: move disk from source to dest
Step3: Hanoi(disk  1, aux, dest, source)
Recurrence relation is: T(n)=2T(n1)+C
Time Complexity: O(2^{n})
→ Tree traversals like Preorder,Postorder and Inorder using recursion only
Note: Recursion is best in the case of Tower of Hanoi,Factorial number and Tree traversals but it is not give good performance to fibonacci numbers.
Question 4 
Domain and Range of the function
Y = √(2x + 3) is
x ≥3/2, y ≥ 0  
x >3/2, y ≤ 0  
x ≥3/2, y ≤ 0  
x ≤3/2, y ≤ 0 
Question 5 
Maximum number of edges in a nNode undirected graph without self loop is
n^{2}  
n(n1)  
n(n+1)  
n(n1)/2 
Question 5 Explanation:
The set of vertices has size n, the number of such subsets is given by the binomial coefficient C(n,2)⋅C(n,2) = n(n1)/2
Question 6 
A hash table has space for 75 records, then the probability of collision before the table is 6% full.
25  
20  
35  
30  
None of the above 
Question 6 Explanation:
Given data,
 Hash table has space =75 slots.
 For 6% filling it must take =6 slots.
 Probability of no collision for first 6 entries=?
Step1: According to given data,
= (^{75}P_{6) / (756) = 0.814586387 Step2: We have to find at least one collision occurs in 6 entries =1Probability of no collision for first 6 entries = 10.814586387 = 0.185413613 Note: Given options are wrong. Given optionB is correct answer according to original key. }
 Hash table has space =75 slots.
 For 6% filling it must take =6 slots.
 Probability of no collision for first 6 entries=?
Step1: According to given data,
= (^{75}P_{6) / (756) = 0.814586387 Step2: We have to find at least one collision occurs in 6 entries =1Probability of no collision for first 6 entries = 10.814586387 = 0.185413613 Note: Given options are wrong. Given optionB is correct answer according to original key. }
Question 7 
BCC in the internet refers to
Black carbon copy  
Blind carbon copy  
Blank carbon copy  
Beautiful carbon copy 
Question 7 Explanation:
→ Blind carbon copy(BCC) is a useful way to let others see an email you sent without the main recipient knowing. It is faster than sending the original message and then forwarding the sent message to the other recipients. It is also good netiquette to use Bcc when copying a message to many people.
Question 8 
Hub is a term used with
A Star Networks  
A Ring Networks  
A Router  
A Bridge 
Question 8 Explanation:
→ Hub is a term used with a star networks. When compare to switch, it is not an intelligent device.
Question 9 
The amount of uncertainty in a system of symbol is called
Bandwidth  
Entropy  
Loss  
Quantum 
Question 9 Explanation:
→ The amount of uncertainty in a system of symbol is called entropy.
→ Generally we can call entropy is disorder or uncertainty.
Question 10 
Which of the following network access standard disassembler is used for connection station to a packet switched network ?
X.3  
X.21  
X.25  
X.75 
Question 10 Explanation:
An X.25 WAN consists of packetswitching exchange (PSE) nodes as the networking hardware, and leased lines, plain old telephone service connections, or ISDN connections as physical links.
Question 11 
A station in a network in a network forward incoming packets by placing them on its shortest output queue. What routing algorithm is being used ?
Hot potato routing  
Flooding  
Static routing  
Delta routing 
Question 11 Explanation:
→ Hotpotato routing is the practice of passing traffic off to another autonomous system as quickly as possible, thus using their network for widearea transit.
→ Coldpotato routing is the opposite, where the originating autonomous system holds onto the packet until it is as near to the destination as possible.
Question 12 
Start and stop bits are used in serial communications for
Error detection
 
Error correction  
Synchronization  
Slowing down the communication 
Question 12 Explanation:
Start and stop bits are used in serial communications for Synchronization.
Question 13 
For a data entry project for office staff who have never used computers before (user interface and user friendliness are extremely important), one will use
Spiral model  
Component based model  
Prototyping  
Waterfall model 
Question 14 
An SRS
establishes the basis for agreement between client and the supplier.  
provides a reference for validation of the final product.  
is a prerequisite to high quality software.  
all of the above. 
Question 14 Explanation:
→ Software Requirement Specification(SRS) establishes the basis for agreement between client and the supplier. It provides a reference for validation of the final product. SRS is a prerequisite to high quality software.
→ The system is considered as a black box whose internal details are not known that is, only its visible external (input/output) behavior is documented.
→ The Software Requirement Specification(SRS) is said to be Consistent if and only if no subset of individual requirements described in it conflict with each other.
→ The system is considered as a black box whose internal details are not known that is, only its visible external (input/output) behavior is documented.
→ The Software Requirement Specification(SRS) is said to be Consistent if and only if no subset of individual requirements described in it conflict with each other.
Question 15 
McCabe’s cyclomatic metric V(G) of a graph G with n vertices, e edges and p connected component ise
e  
n  
e – n + p  
e – n + 2p 
Question 15 Explanation:
Cyclomatic complexity uses 3 formulas
1. Number of regions + 1
2. Predicate + 1
3.EdgesVertices+2(Connected components)
As per the above question, c=cyclomatic complexity
e=number of edges
n=number of vertices
p=predicates
1. Number of regions + 1
2. Predicate + 1
3.EdgesVertices+2(Connected components)
As per the above question, c=cyclomatic complexity
e=number of edges
n=number of vertices
p=predicates
Question 16 
Emergency fixes known as patches are result of
adaptive maintenance  
perfective maintenance  
corrective maintenance  
none of the above 
Question 16 Explanation:
Software maintenance is the modification of a software product after delivery to correct faults, to improve performance or other attributes.
Software maintenance types: 1. Adaptive: modifying the system to cope with changes in the software environment (DBMS, OS)
2. Perfective: Implementing new or changed user requirements which concern functional enhancements to the software.
3. Corrective: Diagnosing and fixing errors, possibly ones found by users.
4. Preventive: Increasing software maintainability or reliability to prevent problems in the future.
Software maintenance types: 1. Adaptive: modifying the system to cope with changes in the software environment (DBMS, OS)
2. Perfective: Implementing new or changed user requirements which concern functional enhancements to the software.
3. Corrective: Diagnosing and fixing errors, possibly ones found by users.
4. Preventive: Increasing software maintainability or reliability to prevent problems in the future.
Question 17 
Design recovery from source code is done during
reverse engineering  
reengineering  
reuse  
all of the above 
Question 17 Explanation:
Design recovery from source code is done during reverse engineering, reengineering and reusability.
Question 18 
Following is used to demonstrate that the new release of software still performs the old one did by rerunning the old tests :
Functional testing  
Path testing  
Stress testing  
Regression testing 
Question 18 Explanation:
→ Regression testing is rerunning functional and nonfunctional tests to ensure that previously developed and tested software still performs after a change. If not, that would be called a regression.
→ Changes that may require regression testing include bug fixes, software enhancements, configuration changes, and even substitution of electronic components.
→ As regression test suites tend to grow with each found defect, test automation is frequently involved.
→ Changes that may require regression testing include bug fixes, software enhancements, configuration changes, and even substitution of electronic components.
→ As regression test suites tend to grow with each found defect, test automation is frequently involved.
Question 19 
The post order traversal of a binary tree is DEBFCA. Find out the preorder traversal.
ABFCDE  
ADBFEC  
ABDECF  
ABDCEF 
Question 19 Explanation:
Post order traversal tree is look like
Postorder: DEBFCA
Inorder: DBEAFC
Preorder: ABDECF
Postorder: DEBFCA
Inorder: DBEAFC
Preorder: ABDECF
Question 20 
B+ tree are preferred to binary tree in database because
Disk capacities are greater than memory capacities  
Disk access much slower than memory access  
Disk data transfer rates are much less than memory data transfer rate  
Disk are more reliable than memory 
Question 20 Explanation:
In B+ trees it is easy to reduce the number of last level access from the disk when the disk size is too large.
Question 21 
What deletes the entire file except the file structure ?
ERASE  
DELETE  
ZAP  
PACK 
Question 21 Explanation:
→ PACK statement removes all records marked for deletion permanently.
→ DELETE statement is used to delete existing records in a table.
→ ZAP deletes the entire file except the file structure.
→ DELETE statement is used to delete existing records in a table.
→ ZAP deletes the entire file except the file structure.
Question 22 
Which command classes text file, which has been created using“SET ALTERNATIVE” <FILE NAME>“Command” ?
SET ALTERNATE OFF  
CLOSE DATABASE  
CLOSE ALTERNATE  
CLEAR ALL 
Question 22 Explanation:
SET ALTERNATE: Controls the recording of input and output in an alternate text file.
Syntax:
SET ALTERNATE on  OFF
SET ALTERNATE TO [  ?  [ADDITIVE]]
→ The default for SET ALTERNATE is OFF. To change the default, set the ALTERNATE parameter in the [OnOffSetting Settings] section of PLUS.ini. To set a default file name for use with SET ALTERNATE, specify an ALTERNATE parameter in the [Command Settings] section of PLUS.ini.
→ Issuing SET ALTERNATE OFF does not close the alternate file. Before accessing the contents of an alternate file, formally close it with CLOSE ALTERNATE or SET ALTERNATE TO (with no file name). This ensures that all data recorded by dBASE Plus for storage in the alternate file is transferred to disk, and automatically turns SET ALTERNATE to OFF.
Syntax:
SET ALTERNATE on  OFF
SET ALTERNATE TO [
→ The default for SET ALTERNATE is OFF. To change the default, set the ALTERNATE parameter in the [OnOffSetting Settings] section of PLUS.ini. To set a default file name for use with SET ALTERNATE, specify an ALTERNATE parameter in the [Command Settings] section of PLUS.ini.
→ Issuing SET ALTERNATE OFF does not close the alternate file. Before accessing the contents of an alternate file, formally close it with CLOSE ALTERNATE or SET ALTERNATE TO (with no file name). This ensures that all data recorded by dBASE Plus for storage in the alternate file is transferred to disk, and automatically turns SET ALTERNATE to OFF.
Question 23 
Data security threats include
privacy invasion  
hardware failure  
fraudulent manipulation of data  
encryption and decryption 
Question 23 Explanation:
→ Hardware failure, encryption and decryption is not related to data security threats.
→ Privacy invasion and fraudulent manipulation of data is related to data security threats. But fraudulent manipulation of data is most appropriate answer.
→ Privacy invasion and fraudulent manipulation of data is related to data security threats. But fraudulent manipulation of data is most appropriate answer.
Question 24 
Which of the following statements is true, when structure of database file with 20 records is modified ?
? EOF( ) Prints. T  
? BOF( ) Prints F  
? BOF( ) Prints T  
? EOF( ) Prints F 
Question 24 Explanation:
Here, EOF means End of File. When structure of database file with 20 records is modified then ?EOF will prints value is T. Here, T means TRUE.
Question 25 
The SQL Expression
Select distinct T. branch name from branch T, branch S where T. assets > S. assets and S. branchcity = DELHI, finds the name of
Select distinct T. branch name from branch T, branch S where T. assets > S. assets and S. branchcity = DELHI, finds the name of
all branches that have greater asset than any branch located in DELHI.  
all branches that have greater assets than allocated in DELHI.  
the branch that has the greatest asset in DELHI.  
any branch that has greater asset than any branch located in DELHI. 
Question 25 Explanation:
Since the query is corelated that is same table is being used two times for comparison so compare each entry of table T with every entry of table S. Then the outcome for above query will be
So the output contains all branches that have greater asset than any branch located in DELHI.
Question 26 
Dijkstra's banking algorithm in an operating system, solves the problem of
deadlock avoidance  
deadlock recovery  
mutual exclusion  
context switching 
Question 26 Explanation:
Dijkstra's banking algorithm in an operating system, solves the problem of deadlock avoidance.
Question 27 
The multiuser operating system, 20 requests are made to use a particular resource per hour, on an average the probability that no request are made in 45 minutes is
e^{–15 }  
e^{–5}  
1 – e^{–5 }  
1 – e^{–10} 
Question 27 Explanation:
Probability of events for a Poisson distribution:
An event can occur 0, 1, 2, … times in an interval. The average number of events in an interval is designated (lambda). is the event rate, also called the rate parameter. The probability of observing k events in an interval is given by the equation
P(k events in interval)=e^{λ} [(λ^{k})/k!]
Where
→ λ is the average number of events per interval
→ e is the number 2.71828... (Euler's number) the base of the natural logarithms
→ k takes values 0, 1, 2, …
→ k! = k × (k − 1) × (k − 2) × … × 2 × 1 is the factorial of k.
→ Now it is given that in 1 hour (60 minutes) 20 requests are made
→ In 1 minute number of requests made= 20/60
→ In 45 minutes total number of requests made = (20/60) * 45
= 15
Probability= (e^{15} *(15)^{0}) / 0!
= e^{15}
An event can occur 0, 1, 2, … times in an interval. The average number of events in an interval is designated (lambda). is the event rate, also called the rate parameter. The probability of observing k events in an interval is given by the equation
P(k events in interval)=e^{λ} [(λ^{k})/k!]
Where
→ λ is the average number of events per interval
→ e is the number 2.71828... (Euler's number) the base of the natural logarithms
→ k takes values 0, 1, 2, …
→ k! = k × (k − 1) × (k − 2) × … × 2 × 1 is the factorial of k.
→ Now it is given that in 1 hour (60 minutes) 20 requests are made
→ In 1 minute number of requests made= 20/60
→ In 45 minutes total number of requests made = (20/60) * 45
= 15
Probability= (e^{15} *(15)^{0}) / 0!
= e^{15}
Question 28 
On receiving an interrupt from an I/O device, the CPU
halts for predetermined time.  
branches off to the interrupt service routine after completion of the current instruction.  
branches off to the interrupt service routine immediately.  
hands over control of address bus and data bus to the interrupting device. 
Question 29 
The maximum amount of information that is available in one portion of the disk access arm for a removal disk pack (without further movement of the arm with multiple heads)
a plate of data  
a cylinder of data  
a track of data  
a block of data 
Question 29 Explanation:
A cylinder of data having maximum amount of information that is available in one portion of the disk access arm for a removal disk pack (without further movement of the arm with multiple heads).
Question 30 
Consider a logical address space of 8 pages of 1024 words mapped with memory of 32 frames. How many bits are there in the physical address ?
9 bits  
11 bits  
13 bits  
15 bits 
Question 30 Explanation:
Since we know page and frame both have the same size and since page size is given as 1024 words, it means frame size is of 1024 i.e. 2^{10}. Hence to uniquely identify each word inside a frame 10bits are needed.
→ Number of frames is given as 32 i.e 2^{5}. So each frame can be uniquely identified using 5bits.
→ Hence the total number of bits needed to identify a word inside memory is 5+10 = 15 bits.
→ Number of frames is given as 32 i.e 2^{5}. So each frame can be uniquely identified using 5bits.
→ Hence the total number of bits needed to identify a word inside memory is 5+10 = 15 bits.
Question 31 
CPU does not perform the operation
data transfer  
logic operation  
arithmetic operation  
all of the above

Question 31 Explanation:
A central processing unit (CPU) is the electronic circuitry within a computer that carries out the instructions of a computer program by performing the basic arithmetic, logical, control and input/output (I/O) operations specified by the instructions.
Question 32 
A chip having 150 gates will be classified as
SSI  
MSI  
LSI  
VLSI 
Question 32 Explanation:
Question 33 
If an integer needs two bytes of storage, then the maximum value of unsigned integer is
2^{16} – 1  
2^{15} – 1  
2^{16}  
2^{15} 
Question 33 Explanation:
= If an integer needs two bytes of storage, then the maximum value of a signed integer is 2^{n1} 1
= 2^{15} 1
= 2^{15} 1
Question 34 
Negative numbers cannot be represented in
signed magnitude form  
1’s complement form  
2’s complement form  
none of the above 
Question 34 Explanation:
Negative numbers can be represented in
1. Signed magnitude form
2. 1’s complement form
3. 2’s complement form
1. Signed magnitude form
2. 1’s complement form
3. 2’s complement form
Question 35 
The cellular frequency reuse factor for the cluster size N is
N  
N^{2}  
1/N  
1/N^{2} 
Question 35 Explanation:
The frequency reuse factor is related to cluster size then the frequency reuse factor=1/N
Question 36 
X – = Y + 1 means
X = X – Y + 1  
X = –X – Y – 1  
X = –X + Y + 1  
= X – Y – 1 
Question 36 Explanation:
In Programming languages like C,C++ and Java we can write X=Y+1 into X=XY+1
Note: X*=Z+1 → X=X*Z+1
X/=Z+1 → X=X/Z+1
X+=Z+1 → X=X+Z+1
Note: X*=Z+1 → X=X*Z+1
X/=Z+1 → X=X/Z+1
X+=Z+1 → X=X+Z+1
Question 37 
Handoff is the mechanism that
transfer an ongoing call from one base station to another  
initiating a new call  
dropping an ongoing call  
none of above 
Question 37 Explanation:
Handoff is the mechanism that transfer an ongoing call from one base station to another.
→ Hard handover (or) Hard Handoff: Early systems used hard handoff. In a hard handoff, a mobile station only communicates with one base station. When the MS moves from one cell to another, communication must first be broken with the previous base station before communication can be established with the new one.
→ Soft handover (or) Soft Handoff: New systems use a soft Handoff. In this case, a mobile station can communicate with base stations at the same time. This means that, during handoff, a mobile station may continue with the new base station before breaking off from the old one.
→ Hard handover (or) Hard Handoff: Early systems used hard handoff. In a hard handoff, a mobile station only communicates with one base station. When the MS moves from one cell to another, communication must first be broken with the previous base station before communication can be established with the new one.
→ Soft handover (or) Soft Handoff: New systems use a soft Handoff. In this case, a mobile station can communicate with base stations at the same time. This means that, during handoff, a mobile station may continue with the new base station before breaking off from the old one.
Question 38 
Which one of the following statement is false ?
Contextfree languages are closed under union.  
Contextfree languages are closed under concatenation.  
Contextfree languages are closed under intersection.  
Contextfree languages are closed under Kleene closure 
Question 38 Explanation:
Contextfree languages are closed under Union,Concatenation and Kleene closure.
Contextfree languages are not closed under set difference,Complementation and intersection.
Contextfree languages are not closed under set difference,Complementation and intersection.
Question 39 
All of the following are examples of real security and privacy risks except
Hackers  
Spam  
Viruses  
Identify theft 
Question 39 Explanation:
→ Hackers,Viruses and identity theft are real security and privacy risks.
→ Spam is not related to real security and privacy risk.
Note: Spam is nothing but unwanted message(or) mail.
→ Spam is not related to real security and privacy risk.
Note: Spam is nothing but unwanted message(or) mail.
Question 40 
Identify the incorrect statement :
The ATM adaptation layer is not service dependent.  
Logical connections in ATM are referred to as virtual channel connections.  
ATM is streamlined protocol with minimal error and flow control capabilities  
ATM is also known as cell delays. 
Question 40 Explanation:
FALSE: The ATM adaptation layer is service dependent.
TRUE: Logical connections in ATM are referred to as virtual channel connections.
TRUE: ATM is streamlined protocol with minimal error and flow control capabilities
TRUE: ATM is also known as cell delays.
TRUE: Logical connections in ATM are referred to as virtual channel connections.
TRUE: ATM is streamlined protocol with minimal error and flow control capabilities
TRUE: ATM is also known as cell delays.
Question 41 
Software risk estimation involves following two tasks :
Risk magnitude and risk impact  
Risk probability and risk impact  
Risk maintenance and risk impact  
Risk development and risk impact 
Question 41 Explanation:
Software Risk estimation involves risk probability and risk impact.
→ Risk impact factors are
1. Negligible [low]
2. Marginal
3. Critical
4. Catastrophic [High]
→ Risk exposure:
RE=P*C
P=The probability of occurrence for a risk
C=The cost to the project should the risk actually occur
→ Risk impact factors are
1. Negligible [low]
2. Marginal
3. Critical
4. Catastrophic [High]
→ Risk exposure:
RE=P*C
P=The probability of occurrence for a risk
C=The cost to the project should the risk actually occur
Question 42 
The number of bits required for an IPv6 address is
16  
32  
64  
128 
Question 42 Explanation:
→ The number of bits required for an IPv6 address is 128.
→ The number of bits required for an IPv4 address is 32
→ The number of bits required for an IPv4 address is 32
Question 43 
The proposition ~qvp is equivalent to
p → q  
q → p  
p ↔ q  
p ∨ q 
Question 43 Explanation:
Hence OptionB is correct answer
Question 44 
Enterprise Resource Planning (ERP)
has existed for over a decade.  
does not integrate well with the functional areas other than operations.  
is inexpensive to implement.  
automate and integrates the majority of business processes 
Question 44 Explanation:
Enterprise Resource Planning (ERP) is automate and integrates the majority of business processes. ERP is the integrated management of core business processes, often in realtime and mediated by software and technology.
Question 45 
Which of the following is false concerning Enterprise Resource Planning (ERP) ?
It attempts to automate and integrate the majority of business processes.  
It shares common data and practices across the enterprise.  
It is inexpensive to implement.  
It provides and access information in a realtime environment. 
Question 45 Explanation:
TRUE: It attempts to automate and integrate the majority of business processes.
TRUE: It shares common data and practices across the enterprise.
FALSE: It is expensive to implement.
TRUE: It provides and access information in a realtime environment.
TRUE: It shares common data and practices across the enterprise.
FALSE: It is expensive to implement.
TRUE: It provides and access information in a realtime environment.
Question 46 
To compare, overlay or cross analyze to maps in GIS
both maps must be in digital form  
both maps must be at the same equivalent scale.  
both maps must be on the same coordinate system  
All of the above 
Question 46 Explanation:
All are correct.
Question 47 
Web Mining is not used in which of the following areas ?
Information filtering  
Crime fighting on the internet  
Online transaction processing  
Click stream analysis. 
Question 47 Explanation:
Web Mining is used to filtering information,online transaction processing and click stream analysis.
Question 48 
A telephone conference call is an example of which type of communications ?
same time / same place  
same time / different place  
different time / different place  
different time / same place 
Question 48 Explanation:
A telephone conference call is an example of same time / different place type of communications.
Question 49 
What is the probability of choosing correctly an unknown integer between 0 and 9 with 3 chances ?
963/1000  
973/1000  
983/1000  
953/1000  
None of the above 
Question 49 Explanation:
Probability of getting a number = 1/10
Probability of not getting a number = 1 1/(10) = 9/10
Now probability that correct number is chosen in first chance= 1/10
probability that correct number is chosen in second chance= (9/10)* (1/10)
probability that correct number is chosen in first chance= (9/10)* (9/10) *(1/10)
So total probability = (1/10)+ [(9/10)* (1/10) ]+ [(9/10)* (9/10) *(1/10)]
= (1/10)+(9/100)+(81/1000)
= 271/1000
Note: Excluded for evaluation because given options are wrong
Probability of not getting a number = 1 1/(10) = 9/10
Now probability that correct number is chosen in first chance= 1/10
probability that correct number is chosen in second chance= (9/10)* (1/10)
probability that correct number is chosen in first chance= (9/10)* (9/10) *(1/10)
So total probability = (1/10)+ [(9/10)* (1/10) ]+ [(9/10)* (9/10) *(1/10)]
= (1/10)+(9/100)+(81/1000)
= 271/1000
Note: Excluded for evaluation because given options are wrong
Question 50 
The number of nodes in a complete binary tree of height h (with roots at level 0) is equal to
2^{0} + 2^{1} + ….. 2^{h}  
2^{0} + 2^{1 }+ ….. 2^{h – 1}  
2^{0} + 2^{1} + ….. 2^{h + 1}  
2^{1} + ….. 2^{h + 1} 
Question 50 Explanation:
The number of nodes in a complete binary tree of height h (with roots at level 0) is equal to 2^{0} + 2^{1 }+ ….. 2^{h}. because every phase we are getting 2^{h}.
Suppose, in level3, it have 2^{3}=8 nodes.
Suppose, in level3, it have 2^{3}=8 nodes.
There are 50 questions to complete.