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There are 180 students at a college following a general course in computing - Edexcel - A-Level Maths Statistics - Question 4 - 2010 - Paper 1
Question 4
There are 180 students at a college following a general course in computing. Students on this course can choose to take up to three extra options. 112 take systems ... show full transcript
Worked Solution & Example Answer:There are 180 students at a college following a general course in computing - Edexcel - A-Level Maths Statistics - Question 4 - 2010 - Paper 1
Step 1
Draw a Venn diagram to represent this information.
Answer
To represent the information using a Venn diagram, start by defining three circles:
- Systems Support (S)
- Developing Software (D)
- Networking (N)
Place the given data in the correct sections:
- The number of students who take all three courses is 4.
- For two intersections:
- Systems Support and Developing Software = 35
- Networking and Developing Software = 28
- Systems Support and Networking = 40
From this, we can determine:
- Only Systems Support = 112 - (36 + 40 + 4) = 32
- Only Developing Software = 70 - (36 + 35 + 4) = 29
- Only Networking = 81 - (28 + 40 + 4) = 9
- The remaining count for each section should add up to total 180.
Step 2
Find the probability that this student takes none of the three extra options.
Answer
Total students is 180.
From the Venn diagram, calculate the total who take at least one option:
- Total who take any option = (S + D + N) - (students counted more than once)
- Here, total taking at least one option = 180 - (those taking none) => students taking none = 180 - 112 = 68.
Thus, the probability = ( P(None) = \frac{68}{180} = \frac{34}{90} = \frac{17}{45} ).
Step 3
Step 4
Find the probability that this student takes all three extra options given that they want to become a technician.
Answer
Among those who want to become technicians, identify how many take Systems Support and Networking.
From the intersections, the students who take both are: (the intersection of Systems Support & Networking) = 40.
Of these, only 4 take all three extra options.
The probability is: ( P(All , Three , Options | Technician) = \frac{4}{40} = \frac{1}{10} ).