| Date | November 2013 | Marks available | 5 | Reference code | 13N.2.sl.TZ0.10 |
| Level | SL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 10 | Adapted from | N/A |
Question
Samantha goes to school five days a week. When it rains, the probability that she goes to school by bus is 0.5. When it does not rain, the probability that she goes to school by bus is 0.3. The probability that it rains on any given day is 0.2.
On a randomly selected school day, find the probability that Samantha goes to school by bus.
Given that Samantha went to school by bus on Monday, find the probability that it was raining.
In a randomly chosen school week, find the probability that Samantha goes to school by bus on exactly three days.
After \(n\) school days, the probability that Samantha goes to school by bus at least once is greater than \(0.95\). Find the smallest value of \(n\).
Markscheme
appropriate approach (M1)
eg \({\text{P}}(R \cap B) + {\text{P}}(R' \cap B)\), tree diagram,
one correct multiplication (A1)
eg \(0.2 \times 0.5,{\text{ }}0.24\)
correct working (A1)
eg \(0.2 \times 0.5 + 0.8 \times 0.3,{\text{ }}0.1 + 0.24\)
\({\text{P(bus)}} = 0.34 {\text{(exact)}}\) A1 N3
[4 marks]
recognizing conditional probability (R1)
eg \({\text{P}}(A|B) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}}\)
correct working A1
eg \(\frac{{0.2 \times 0.5}}{{0.34}}\)
\({\text{P}}(R|B) = \frac{5}{{17}},{\text{ }}0.294\) A1 N2
[3 marks]
recognizing binomial probability (R1)
eg \(X \sim {\text{B}}(n,{\text{ }}p)\), \(\left( \begin{array}{c}5\\3\end{array} \right)\) \({(0.34)^3},{\text{ }}{(0.34)^3}{(1 - 0.34)^2}\)
\({\text{P}}(X = 3) = 0.171\) A1 N2
[2 marks]
METHOD 1
evidence of using complement (seen anywhere) (M1)
eg \(1 - {\text{P (none), }}1 - 0.95\)
valid approach (M1)
eg \(1 - {\text{P (none)}} > 0.95,{\text{ P (none)}} < 0.05,{\text{ }}1 - {\text{P (none)}} = 0.95\)
correct inequality (accept equation) A1
eg \(1 - {(0.66)^n} > 0.95,{\text{ }}{(0.66)^n} = 0.05\)
\(n > 7.209{\text{ (accept }}n = 7.209{\text{)}}\) (A1)
\(n = 8\) A1 N3
METHOD 2
valid approach using guess and check/trial and error (M1)
eg finding \({\text{P}}(X \geqslant 1)\) for various values of n
seeing the “cross over” values for the probabilities A1A1
\(n = 7,{\text{ P}}(X \geqslant 1) = 0.9454,{\text{ }}n = 8,{\text{ P}}(X \geqslant 1) = 0.939\)
recognising \(0.9639 > 0.95\) (R1)
\(n = 8\) A1 N3
[5 marks]
