Question: a community chief wants to organize training for the community members that use energy sources that do not conserve the environment to cook .He carre=ied out a survey on a sample of 130 community members and discovered that :62 use firewood ,25 use firewood and gas ,28 use firewood and electricity .Those who use electricity but not other sources are are 12 .Those who use gas only are more than those who use firewood only by 5. Those who use firewood or gas exclusive of electricity were 60. Those that use gas and electricity were 29.Some used all and some used none .The training will be organised if the pronanility of those using only surch resources is greater than 0.2. SORT THE DATA INTO USING ONE ONLY TWO ALL THREE AND NONE OF THE SOURCES IN TO A SET OF THREE CIRCLES

1. **State the problem:** We have a community of $130$ members using three energy sources: firewood (F), gas (G), and electricity (E). We want to classify members into groups using only one source, exactly two sources, all three sources, or none, and check if the probability of those using only such sources is greater than $0.2$. 2. **Given data:** - Total members $=130$ - $|F|=62$ - $|F \cap G|=25$ - $|F \cap E|=28$ - $|E \text{ only}|=12$ - $|G \text{ only}| = |F \text{ only}| + 5$ - $|F \text{ or } G \text{ exclusive of } E|=60$ - $|G \cap E|=29$ - Some use all three, some use none. 3. **Define variables:** Let: - $x = |F \text{ only}|$ - $y = |G \text{ only}| = x + 5$ - $z = |E \text{ only}| = 12$ - $a = |F \cap G \cap E|$ (all three) - $b = |F \cap G| - a = 25 - a$ - $c = |F \cap E| - a = 28 - a$ - $d = |G \cap E| - a = 29 - a$ 4. **Use the condition for $F$ or $G$ exclusive of $E$:** This means members who use $F$ or $G$ but not $E$: $$|F \text{ only}| + |G \text{ only}| + |F \cap G| - |F \cap G \cap E| = 60$$ Substitute: $$x + y + b = 60$$ $$x + (x+5) + (25 - a) = 60$$ $$2x + 30 - a = 60$$ $$2x - a = 30$$ 5. **Use total for $F$:** $$|F| = |F \text{ only}| + |F \cap G| + |F \cap E| - |F \cap G \cap E|$$ $$62 = x + 25 + 28 - a$$ $$62 = x + 53 - a$$ $$x - a = 9$$ 6. **Solve the system:** From step 4: $$2x - a = 30$$ From step 5: $$x - a = 9$$ Subtract second from first: $$2x - a - (x - a) = 30 - 9$$ $$x = 21$$ Then from step 5: $$21 - a = 9 \Rightarrow a = 12$$ 7. **Calculate other values:** $$y = x + 5 = 21 + 5 = 26$$ $$b = 25 - a = 25 - 12 = 13$$ $$c = 28 - a = 28 - 12 = 16$$ $$d = 29 - a = 29 - 12 = 17$$ 8. **Calculate total members accounted for:** Sum all disjoint groups: - Only one source: $x + y + z = 21 + 26 + 12 = 59$ - Exactly two sources (excluding all three): $b + c + d = 13 + 16 + 17 = 46$ - All three: $a = 12$ Total so far: $$59 + 46 + 12 = 117$$ 9. **Calculate those using none:** $$130 - 117 = 13$$ 10. **Check probability of using only one source:** $$P = \frac{59}{130} \approx 0.4538 > 0.2$$ 11. **Conclusion:** Training will be organized since the probability of using only one source is greater than $0.2$. **Final classification:** - Only Firewood: $21$ - Only Gas: $26$ - Only Electricity: $12$ - Exactly two sources: $46$ - All three sources: $12$ - None: $13$