Wireshark packet capture basics in Cybersecurity - Practice Problems & Coding Challenges

Challenge - 5 Problems

📋 Factual

intermediate

What is the primary function of Wireshark?

Wireshark is a popular tool used in cybersecurity. What is its main purpose?

A To capture and analyze network packets in real time

B To encrypt network traffic for secure communication

C To block unauthorized network access

D To scan computers for viruses and malware

💡 Hint

Think about what Wireshark does with data moving through a network.

Wireshark captures network packets and allows users to inspect their contents to understand network behavior and troubleshoot issues.

🧠 Conceptual

intermediate

Which filter would you use to see only HTTP traffic in Wireshark?

You want to view only web traffic using Wireshark. Which display filter should you apply?

A tcp.port == 80

B udp.port == 53

C icmp

D tcp.port == 443

💡 Hint

HTTP usually uses a specific TCP port number.

HTTP traffic typically uses TCP port 80, so filtering by tcp.port == 80 shows only HTTP packets.

🔍 Analysis

advanced

What does the 'No. of packets' column represent in Wireshark capture summary?

In Wireshark's capture summary, what does the 'No. of packets' column indicate?

A The number of packets sent by the selected device only

B The total number of packets captured in the entire session

C The number of packets in the current filtered view

D The number of packets dropped during capture

💡 Hint

Consider what the summary shows before applying any filters.

The 'No. of packets' column shows the total packets captured during the session, regardless of filters.

❓ Reasoning

advanced

Why might some packets be missing in a Wireshark capture?

During a capture, you notice some expected packets are missing. What is a likely reason?

A Wireshark automatically deletes packets after 10 seconds

B The network cable was unplugged during capture

C The packets were encrypted and cannot be captured

D The capture interface was not set to promiscuous mode

💡 Hint

Think about how network interfaces receive packets not addressed to them.

If the interface is not in promiscuous mode, it only captures packets addressed to it, missing others on the network.

❓ Comparison

expert

Which statement correctly compares capture filters and display filters in Wireshark?

Choose the option that accurately describes the difference between capture filters and display filters in Wireshark.

A Capture filters are applied after capture; display filters are applied during capture

B Display filters limit which packets are saved during capture; capture filters only hide packets after capture

C Capture filters limit which packets are saved during capture; display filters only hide packets after capture

D Both capture and display filters work the same way and can be used interchangeably

💡 Hint

Consider when each filter type is applied in the capture process.

Capture filters decide which packets are recorded during capture, reducing file size. Display filters only change what is shown after capture without removing packets.

Practice

1. What is the primary purpose of Wireshark in cybersecurity?

easy

A. To capture and analyze network packets in real time

B. To encrypt network traffic for security

C. To block unauthorized network access

D. To create virtual private networks (VPNs)

Solution

1. Step 1: Understand Wireshark's function

   Wireshark is a tool designed to capture and display network packets as they travel through a network.

2. Step 2: Identify the correct purpose

   Among the options, only capturing and analyzing packets matches Wireshark's main use.

3. Final Answer:

   To capture and analyze network packets in real time -> Option A

4. Quick Check:

   Wireshark captures packets = To capture and analyze network packets in real time [OK]

Hint: Wireshark shows network data live, not encrypt or block [OK]

Common Mistakes:

• Confusing Wireshark with firewall or VPN tools
• Thinking Wireshark encrypts data
• Assuming Wireshark blocks traffic

2. Which of the following is the correct way to start a packet capture in Wireshark?

easy

A. Click on 'File' then 'Open Capture'

B. Click on 'Capture' then 'Start'

C. Click on 'Analyze' then 'Filter'

D. Click on 'Edit' then 'Preferences'

Solution

1. Step 1: Identify the menu for starting capture

   In Wireshark, the 'Capture' menu contains options to start or stop capturing packets.

2. Step 2: Match the correct action

   Clicking 'Capture' then 'Start' begins the live packet capture process.

3. Final Answer:

   Click on 'Capture' then 'Start' -> Option B

4. Quick Check:

   Start capture via Capture menu = Click on 'Capture' then 'Start' [OK]

Hint: Start capture under 'Capture' menu, not 'File' or 'Edit' [OK]

Common Mistakes:

• Choosing 'File' to start capture instead of 'Capture'
• Confusing 'Analyze' with starting capture
• Looking in 'Edit' menu for capture options

3. Consider the following Wireshark filter: ip.src == 192.168.1.10. What does this filter do?

medium

A. Shows packets where the destination IP is 192.168.1.10

B. Shows packets with any IP address except 192.168.1.10

C. Shows packets where the source IP is 192.168.1.10

D. Shows packets where either source or destination IP is 192.168.1.10

Solution

1. Step 1: Understand the filter syntax

   The filter ip.src == 192.168.1.10 means packets where the source IP address equals 192.168.1.10.

2. Step 2: Match filter meaning to options

   Only Shows packets where the source IP is 192.168.1.10 correctly describes packets with source IP 192.168.1.10.

3. Final Answer:

   Shows packets where the source IP is 192.168.1.10 -> Option C

4. Quick Check:

   ip.src filter = source IP = Shows packets where the source IP is 192.168.1.10 [OK]

Hint: ip.src means source IP, ip.dst means destination IP [OK]

Common Mistakes:

• Confusing source IP with destination IP
• Assuming filter matches both source and destination
• Thinking filter excludes the IP address

4. You tried to filter packets with tcp.port == 80 but no packets appear. What could be a likely reason?

medium

A. Port 80 is not used for TCP traffic

B. The filter syntax is incorrect

C. Wireshark does not support filtering by port

D. You captured packets on the wrong network interface

Solution

1. Step 1: Check filter syntax correctness

   The filter tcp.port == 80 is valid syntax to filter TCP packets on port 80.

2. Step 2: Consider capture context

   If no packets appear, a common cause is capturing on the wrong network interface where no HTTP traffic (port 80) passes.

3. Final Answer:

   You captured packets on the wrong network interface -> Option D

4. Quick Check:

   Wrong interface capture = no matching packets = You captured packets on the wrong network interface [OK]

Hint: No packets? Check if capturing on correct network interface [OK]

Common Mistakes:

• Assuming filter syntax is wrong without checking
• Believing Wireshark can't filter by port
• Thinking port 80 is not TCP by default

5. You want to capture only HTTP traffic from a specific device with IP 10.0.0.5 using Wireshark. Which filter should you apply?

hard

A. ip.addr == 10.0.0.5 and tcp.port == 80

B. ip.src == 10.0.0.5 or ip.dst == 10.0.0.5 and tcp.port == 80

C. ip.addr == 10.0.0.5 or tcp.port == 80

D. ip.src == 10.0.0.5 and tcp.port == 80

Solution

1. Step 1: Define the filter requirements

   You want packets where the device IP is either source or destination and the traffic is HTTP (TCP port 80).

2. Step 2: Analyze each filter option

   ip.addr == 10.0.0.5 and tcp.port == 80 uses ip.addr == 10.0.0.5 which matches source or destination IP, combined with tcp.port == 80 to filter HTTP traffic. This matches the requirement exactly.

3. Step 3: Identify issues in other options

   ip.src == 10.0.0.5 or ip.dst == 10.0.0.5 and tcp.port == 80 lacks parentheses, so 'or' and 'and' precedence causes incorrect filtering. ip.addr == 10.0.0.5 or tcp.port == 80 matches any packet with IP 10.0.0.5 or any TCP port 80 packet, which is too broad. ip.src == 10.0.0.5 and tcp.port == 80 only matches packets where 10.0.0.5 is source, missing destination packets.

4. Final Answer:

   ip.addr == 10.0.0.5 and tcp.port == 80 -> Option A

5. Quick Check:

   ip.addr covers both ends + tcp.port 80 = ip.addr == 10.0.0.5 and tcp.port == 80 [OK]

Hint: Use ip.addr for both source/destination IPs in filters [OK]

Common Mistakes:

• Not using parentheses causing wrong logic in filters
• Using only ip.src or ip.dst missing half the traffic
• Using 'or' instead of 'and' causing too many packets