Four persons can hit a target correctly with probabilities frac{1}{2},frac{1}{3},frac{1}{4} an

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14.Probability

hard

Four persons can hit a target correctly with probabilities $\frac{1}{2},\frac{1}{3},\frac{1}{4}$ and $\frac {1}{8}$ respectively. If all hit at the target independently, then the probability that the target would be hit, is

$\frac{{25}}{{32}}$

$\frac{{25}}{{192}}$

$\frac{{7}}{{32}}$

$\frac{{1}}{{192}}$

(JEE MAIN-2019)

Solution

Let persons be $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$

$P(H i t)=1-P$ (none of them hits)

$=1-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}} \cap \overline{\mathrm{D}})$

$=1-\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\overline{\mathrm{B}}) \cdot \mathrm{P}(\overline{\mathrm{C}}) \cdot \mathrm{P}(\overline{\mathrm{D}})$

$=1-\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{7}{8}$

$=\frac{25}{32}$

Standard 11

Mathematics

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easy

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easy

Four persons can hit a target correctly with probabilities $\frac{1}{2},\frac{1}{3},\frac{1}{4}$ and $\frac {1}{8}$ respectively. If all hit at the target independently, then the probability that the target would be hit, is

Solution

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