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Question Number 12649 by tawa last updated on 28/Apr/17

Answered by mrW1 last updated on 28/Apr/17

((300+500+700−440−1000)/2)=30  ⇒30 persons involved lost of injury  to all the three.

$$\frac{\mathrm{300}+\mathrm{500}+\mathrm{700}−\mathrm{440}−\mathrm{1000}}{\mathrm{2}}=\mathrm{30} \\ $$$$\Rightarrow\mathrm{30}\:{persons}\:{involved}\:{lost}\:{of}\:{injury} \\ $$$${to}\:{all}\:{the}\:{three}. \\ $$

Commented by tawa last updated on 28/Apr/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by mrW1 last updated on 29/Apr/17

Explanation:  x persons with injury to only one of the three  y persons with injury to two of the three  z persons with injury to all of the three  x+y+z=1000       ...(i)  x+2y+3z=300+500+700     ...(ii)  y=440  ⇒z=((300+500+700−1000−440)/2)=30  ⇒x=1000−440−30=530

$${Explanation}: \\ $$$${x}\:{persons}\:{with}\:{injury}\:{to}\:{only}\:{one}\:{of}\:{the}\:{three} \\ $$$${y}\:{persons}\:{with}\:{injury}\:{to}\:{two}\:{of}\:{the}\:{three} \\ $$$${z}\:{persons}\:{with}\:{injury}\:{to}\:{all}\:{of}\:{the}\:{three} \\ $$$${x}+{y}+{z}=\mathrm{1000}\:\:\:\:\:\:\:...\left({i}\right) \\ $$$${x}+\mathrm{2}{y}+\mathrm{3}{z}=\mathrm{300}+\mathrm{500}+\mathrm{700}\:\:\:\:\:...\left({ii}\right) \\ $$$${y}=\mathrm{440} \\ $$$$\Rightarrow{z}=\frac{\mathrm{300}+\mathrm{500}+\mathrm{700}−\mathrm{1000}−\mathrm{440}}{\mathrm{2}}=\mathrm{30} \\ $$$$\Rightarrow{x}=\mathrm{1000}−\mathrm{440}−\mathrm{30}=\mathrm{530} \\ $$

Commented by tawa last updated on 28/Apr/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by tawa last updated on 28/Apr/17

sir is it z = ((......)/3)  ???   or   ((.......)/2) ???  because you said 3z

$$\mathrm{sir}\:\mathrm{is}\:\mathrm{it}\:\mathrm{z}\:=\:\frac{......}{\mathrm{3}}\:\:???\:\:\:\mathrm{or}\:\:\:\frac{.......}{\mathrm{2}}\:??? \\ $$$$\mathrm{because}\:\mathrm{you}\:\mathrm{said}\:\mathrm{3z} \\ $$

Commented by mrW1 last updated on 28/Apr/17

(ii)−(i) gets  y+2z=300+500+700−1000  2z=300+500+700−1000−440  z=((300+500+700−1000−440)/2)

$$\left({ii}\right)−\left({i}\right)\:{gets} \\ $$$${y}+\mathrm{2}{z}=\mathrm{300}+\mathrm{500}+\mathrm{700}−\mathrm{1000} \\ $$$$\mathrm{2}{z}=\mathrm{300}+\mathrm{500}+\mathrm{700}−\mathrm{1000}−\mathrm{440} \\ $$$${z}=\frac{\mathrm{300}+\mathrm{500}+\mathrm{700}−\mathrm{1000}−\mathrm{440}}{\mathrm{2}} \\ $$

Commented by tawa last updated on 28/Apr/17

i understand now sir. God bless you sir.

$$\mathrm{i}\:\mathrm{understand}\:\mathrm{now}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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