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2005 Paper 1 Q12
D: 1500.0
B: 1529.8
probability
principle of inclusion-exclusion
conditional probability
inequality
Venn diagram
range of values
- The probability that a hobbit smokes a pipe is 0.7 and the probability that a hobbit wears a hat is 0.4. The probability that a hobbit smokes a pipe but does not wear a hat is \(p\). Determine the range of values of \(p\) consistent with this information.
- The probability that a wizard wears a hat is 0.7; the probability that a wizard wears a cloak is 0.8; and the probability that a wizard wears a ring is 0.4. The probability that a wizard does not wear a hat, does not wear a cloak and does not wear a ring is 0.05. The probability that a wizard wears a hat, a cloak and also a ring is 0.1. Determine the probability that a wizard wears exactly two of a hat, a cloak, and a ring. The probability that a wizard wears a hat but not a ring, given that he wears a cloak, is \(q\). Determine the range of values of \(q\) consistent with this information.
Solution:
- \(\,\)
The overlap can be at most 0.4, which would mean \(p =0.7-0.4 = 0.3\) It must be at least 0.1, which would mean \(p =0.7-0.1 = 0.6\) so \(0.3 \leq p \leq 0.6\)
- Notice that: \begin{align*} && 1 &= 0.05 + 0.7 -(hc+hr+0.1) + \\ &&&\quad\quad 0.8 - (hc+cr + 0.1) + \\ &&&\quad \quad \quad 0.4 - (hr+cr+0.1) +\\ &&&\quad \quad \quad \quad hc+hr+cr+0.1 \\ && &= 0.05 +0.7+0.8+0.4 - (hc+hr+cr)-2\cdot 0.1 \\ \Rightarrow && hc+hr+cr &=0.05 +0.7 + 0.8 + 0.4 - 0.2-1 \\ \Rightarrow && \mathbb{P}(\text{exactly 2}) &= 0.75 \end{align*} Notice \(q = \frac{hc}{0.8}\) Notice that we must have: \(hc, hr cr \geq 0\) as well as \(hc+hr+cr = 0.75\) \begin{align*} && \mathbb{P}(\text{only hat}) &= 0.7 -(hc+hr+0.1) \geq 0 \\ \Rightarrow && hc+hr & \leq 0.6 \\ && \mathbb{P}(\text{only cloak}) &= 0.8 - (hc+cr + 0.1)\geq 0 \\ \Rightarrow &&hc+cr & \leq 0.7 \\ && \mathbb{P}(\text{only ring}) &= 0.4 - (hr+cr+0.1) \geq 0 \\ \Rightarrow && hc+hr & \leq 0.3 \\ \end{align*} To find the minimum for \(hc\) we want to maximise \(hr+cr = 0.3\), so \(hc = 0.75 - 0.3 = 0.45\). To find the maximum for \(hc\) we want to minimise \(hr\) and \(cr\) \(cr \leq 0.7 - hc\) and \(hr \leq 0.6 - hc\) so \(0.75 \leq hc + (0.6 - hc) + (0.7 - hc) = 1.3-hc\) so \(hc \leq 1.3 - 0.75 = 0.55\) Therefore the range for \(q\) is \(\frac{.45}{.8}\) to \(\frac{.55}{.8}\) or \(\frac9{16} \leq q \leq \frac{11}{16}\)
