In a battle,70% of the combatants lost one ey | vedclass

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(A) Let the total number of combatants be $100$. Let $A, B, C, D$ be the sets of combatants who lost an eye,an ear,an arm,and a leg,respectively. Give

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) Let the total number of combatants be $100$. Let $A, B, C, D$ be the sets of combatants who lost an eye,an ear,an arm,and a leg,respectively. Given: $n(A) = 70, n(B) = 80, n(C) = 75, n(D) = 85$. The number of people who did $NOT$ lose these parts are: $n(A^c) = 100 - 70 = 30$ $n(B^c) = 100 - 80 = 20$ $n(C^c) = 100 - 75 = 25$ $n(D^c) = 100 - 85 = 15$ The number of people who lost at least one part is $100 - n(A^c \cap B^c \cap C^c \cap D^c)$. To minimize $x$ (the intersection of all four),we maximize the union of the complements. The maximum number of people who lost at least one part is $100$. Using the principle of inclusion-exclusion for the complements: $n(A^c \cup B^c \cup C^c \cup D^c) \leq n(A^c) + n(B^c) + n(C^c) + n(D^c) = 30 + 20 + 25 + 15 = 90$. Thus,the number of people who lost all four parts is at least $100 - 90 = 10$. Therefore,the minimum value of $x$ is $10$.

In a battle,$70\%$ of the combatants lost one eye,$80\%$ an ear,$75\%$ an arm,$85\%$ a leg,and $x\%$ lost all the four limbs. What is the minimum value of $x$?

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