Question involving application of inclusion-exclusion principle

Question

A doctor is conducting a study relating patients blood pressure to the regularity of their heartbeat and has the following statistics:

1. $14$% have high blood pressure ($H$)
2. $22$% have low blood pressure ($L$)
3. $15$% have irregular heartbeat ($I$)
4. $\frac{1}{3}$ of those with $I$ have $H$
5. $\frac{1}{8}$ of those with normal blood pressure $N$ have $I$.

What portion has a regular heartbeat ($R$) and $L$?

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For this I first wanted to establish all the statistical quantities to be derived from arithmetic: To that end we find that:

• $n(R) = 100 - 15 = 85$
• $n(N) = 100 - 22 - 14 = 64$

From here we can establish the rest of the given statistics using set theoretic notation

• $n(H \cap I) = \frac{1}{3} \cdot n(I) = 5$
• $n(N \cap I) = \frac{1}{8} \cdot n(N) = 8$

Now I know we want to find the value of $n(R \cap L)$ so we can invoke the inclusion-exclusion principle for $R$ and $L$ and see that

• $n(R \cup L) = n(R) + n(L) - n(R \cap L) \implies n(R \cap L) = 107 - n(R \cup L)$

From here I wanted to try using De Morgan's Laws regarding the complement of a union of two sets is the intersection of their complements. However, I found that the complement of $N$ would be the union of $L$ and $H$ minus their respective intersections with $N$ but tht would complicate things too much and would lead me in circles. So, I'm hoping someone can give me a hint as to how to proceed from here. I imagine I have to use the knowledge of the size of the intersections of $H$ and $N$ wuth $I$ but am having trouble seeing where that comes in.

Answer 1

You have two criteria:

• blood pressure, with 3 outcomes: $H$, $L$, or $N$;
• heartbeat, with 2 outcomes $I$ or $R$.

By combining the two criteria, the set of patients is partitioned in 6 sets: $HI, LI, NI, HR, LR, NR$.

The data you have says that $P(H)=0.14, P(L)=0.22$, so $P(N)=1-P(H)-P(L)=0.64$. Since $\dfrac{P(NI)}{P(N)}=\dfrac{1}{8}$, then $P(NI)=0.08$.

Also, $P(I)=0.15$, and $\dfrac{P(HI)}{P(I)}=\dfrac{1}{3}$, so $P(HI)=0.05$.

Now, $P(I)=P(LI)+P(NI)+P(RI)$, so $P(LI)=P(I)-P(NI)-P(RI)=0.15-0.08-0.05=0.02$.

Finally, $P(L)=P(LI)+P(LR)$, so $P(LR)=P(L)-P(LI)=0.22-0.02=0.2$, so the answer is $20\%$.

Note on notation: $P$ stands for percent or probability and I omitted the intersection symbol (which is quite standard in probability theory).

Answer 2

Since the universe is disjoint union of $H$, $N$, and $L$, then: $n(I)$ = $n(H \cap I) + n(N \cap I) + n(L \cap I)$.

From your working, you can get $n(L \cap I) = 2$. It's quite straightforward from this to get $n(L \cap R) = 22 - 2 = 20$.

Answer 3

To apply P.I.E. one needs either $R \cap L$ or $R \cup L$.

However, inspecting the table, we see that $R \cap L$ shows up first !

Line I gives $I \cap L = 15-5-8 = 2$

Column L gives $R \cap L = 22-2 = 20$

Now we can apply P.I.E to obtain $R \cup L = 85 + 22 - 20 = 87$

\begin{array}{c|c|c|c} & H = 14 & L = 22 & N=64 \\ \hline R=85 & . &. & . \\ I=15 & 5 & . & 8\\ \hline \end{array}

We can enforce P.I.E. in opposite direction via De Morgan, which says that $87 = 100 - 5 - 8$.