| Marks available | 5 |
| Reference code | 16M.1.sl.TZ2.8 |
Question
In a class of 21 students, 12 own a laptop, 10 own a tablet, and 3 own neither.
The following Venn diagram shows the events “own a laptop” and “own a tablet”.
The values \(p\), \(q\), \(r\) and \(s\) represent numbers of students.
A student is selected at random from the class.
Two students are randomly selected from the class. Let \(L\) be the event a “student owns a laptop”.
(i) Write down the value of \(p\).
(ii) Find the value of \(q\).
(iii) Write down the value of \(r\) and of \(s\).
(i) Write down the probability that this student owns a laptop.
(ii) Find the probability that this student owns a laptop or a tablet but not both.
(i) Copy and complete the following tree diagram. (Do not write on this page.)
(ii) Write down the probability that the second student owns a laptop given that the first owns a laptop.
Markscheme
(i) \(p = 3\) A1 N1
(ii) valid approach (M1)
eg\(\,\,\,\,\,\)\((12 + 10 + 3) – 21,{\text{ }}22 – 18\)
\(q = 4\) A1 N2
(iii) \(r = 8,{\text{ }}s = 6\) A1A1 N2
(i) \(\frac{{12}}{{21}}{\text{ }}\left( { = \frac{4}{7}} \right)\) A2 N2
(ii) valid approach (M1)
eg\(\,\,\,\,\,\)\(8 + 6,{\text{ }}r + s\)
\(\frac{{14}}{{21}}{\text{ }}\left( { = \frac{2}{3}} \right)\) A1 N2
(i) 
A1A1A1 N3
(ii) \(\frac{{11}}{{20}}\) A1 N1
[4 marks]
Examiners report
On the whole, candidates were very successful on this question, with the majority of candidates earning most of the available marks.
On the whole, candidates were very successful on this question, with the majority of candidates earning most of the available marks.
On the whole, candidates were very successful on this question, with the majority of candidates earning most of the available marks. The most common error was seen in part (c)(ii), where many candidates did not earn the mark. It is also interesting to note that many of the candidates who answered this part correctly did so by using the formula for conditional probability, rather than recognizing that the required probability is given to them in the second branch of the tree diagram.
| Marks available | 3 |
| Reference code | 18M.1.sl.TZ2.8 |
Question
Pablo drives to work. The probability that he leaves home before 07:00 is \(\frac{3}{4}\).
If he leaves home before 07:00 the probability he will be late for work is \(\frac{1}{8}\).
If he leaves home at 07:00 or later the probability he will be late for work is \(\frac{5}{8}\).
Copy and complete the following tree diagram.
Find the probability that Pablo leaves home before 07:00 and is late for work.
Find the probability that Pablo is late for work.
Given that Pablo is late for work, find the probability that he left home before 07:00.
Two days next week Pablo will drive to work. Find the probability that he will be late at least once.
Markscheme
A1A1A1 N3
Note: Award A1 for each bold fraction.
[3 marks]
multiplying along correct branches (A1)
eg \(\frac{3}{4} \times \frac{1}{8}\)
P(leaves before 07:00 ∩ late) = \(\frac{3}{32}\) A1 N2
[2 marks]
multiplying along other “late” branch (M1)
eg \(\frac{1}{4} \times \frac{5}{8}\)
adding probabilities of two mutually exclusive late paths (A1)
eg \(\left( {\frac{3}{4} \times \frac{1}{8}} \right) + \left( {\frac{1}{4} \times \frac{5}{8}} \right),\,\,\frac{3}{{32}} + \frac{5}{{32}}\)
\({\text{P}}\left( L \right) = \frac{8}{{32}}\,\,\left( { = \frac{1}{4}} \right)\) A1 N2
[3 marks]
recognizing conditional probability (seen anywhere) (M1)
eg \({\text{P}}\left( {A|B} \right),\,\,{\text{P}}\left( {{\text{before 7}}|{\text{late}}} \right)\)
correct substitution of their values into formula (A1)
eg \(\frac{{\frac{3}{{32}}}}{{\frac{1}{4}}}\)
\({\text{P}}\left( {{\text{left before 07:00}}|{\text{late}}} \right) = \frac{3}{8}\) A1 N2
[3 marks]
valid approach (M1)
eg 1 − P(not late twice), P(late once) + P(late twice)
correct working (A1)
eg \(1 – \left( {\frac{3}{4} \times \frac{3}{4}} \right),\,\,2 \times \frac{1}{4} \times \frac{3}{4} + \frac{1}{4} \times \frac{1}{4}\)
\(\frac{7}{{16}}\) A1 N2
[3 marks]